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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at http : //books . google . com/| d by Google Kci*fTOi Di'ized by Google d by Google d by Google THE LONDON SCIENCE CLASS-BOOKS ELEMENTARY SERIES EDITED BY PROF. G. C. FOSTER, F.R.S. and SIR PHILIP MAGNUS HYDROSTATICS AND PNEUMATICS d by Google d by Google HYDROSTATICS AND PNEUMATICS BY SIR PHILIP MAGNUS OF THE CITY AND GUILDS OF LONDON INSTITUTE ; AUTHOR OF 'lessons IN ELEMENTARY MECHANICS/ &C JOINT EDITOR OF THIS SERIES EIGHTH EDITION LONDON LONGMANS, GREEN, AND CO. AND NEW YORK: 1 5 EAST 1 6* STREET 189I All rights reserved Digitized by VjOOQIC /{c»4lo I HARVARD UNIVERSITY LIBRARY ' ^92 ^ Richard Clay & Sons, Limited, London & Bungay. d by Google EDITORS' PREFACE. Notwithstanding the large number of scientific works which have been published within the last few years, it is very generally acknowledged by those who are practically engaged in Education, whether as Teachers or as Examiners, that there is still a want of Books adapted for school purposes upon several important branches of Science. The present Series will aim at supplying this deficiency. The works comprised in the Series will all be composed with * special reference to their use in school-teaching ; but, at the same time, particular attention will be given to making the information contained in them trust- worthy and accurate, and to presenting it in such a way that it may serve as a basis for more advanced study. In conformity with the special object of the Series, the attempt will be made in all cases to bring out the educational value which properly belongs to the study of any branch of Science, by not merely treating of its d by Google VI Editors' Preface, acquired results, but by explaining as fully as possible the nature of the methods of inquiry and reasoning by which these results have been obtained. Conse- quently, although the treatment of each subject will be strictly elementary, the fundaro'^ntal facts will be stated and discussed with the fulness needed to place their scientific significance in a clear light, and to show the relation in which they stand to the general conclusions of Science. In order to ensure the efficient carrying-out of the general scheme indicated above, the Editors have endeavoured to obtain the co-operation, as Authors of the several treatises, of men who combine special knowledge of the subjects on which they write with practical experience in Teaching. The volumes of the Series will be published, if possible, at a uniform price of \s, 6d, It is intended • that eventually each of the chief branches of Science shall be represented by one or more volumes. G. C. F P. M. d by Google PREFACE. This Class-Book is intended for the use of those pupils in the upper forms of schools who have already acquired some elementary knowledge of the principles of Mechanics, and are about to commence the study of Hydrostatics and Pneumatics. In the treatment of the subject of this volume I have endeavoured, as far as possible, to combine the Experimental with the Deductive method. When- ever a law is stated, some explanation is afforded of the several experiments by which that law has been established ; and whenever a result is deduced, by the aid of mathematical reasoning, from more ele- mentary principles, the pupil is shown how this result may be experimentally verified In the hope that this Uttle work may serve as an introduction to more advanced treatises on Hydro- statics, I have devoted a few pages to the consider- ation of the * Flow of Liquids through Pipes and Small Orifices'; and, whilst avoiding the mathe- d by Google yiii Preface. matical difficulties which the fuller treatment of this branch of the subject involves, I have endeavoured to bring into prominence some of the leading prin- ciples connected with it, which recent investigations have aimed at establishing. To facilitate the use of this text-book in class- instruction, the subject-matter is divided into a number, of short sections, in which all the more im- portant propositions are illustrated by numerical examples. To nearly every section is appended a set of exercises, progressively arranged, to be solved by the pupil. My obligations to the published works of different writers are acknowledged in the body of the book. P. M London^ Savile Club : September 1878. IMs volume can be obtained with or mthout the answers to the Exerciser d by Google CONTENTS. CHAPTER I. PRELIMINARY NOTIOKS. '.ECTION PAGR I. Nature of Fluid Bodies I Essential Diflferences between Solid and Fluid Bodies. — Change of Gaseous into Liquid State. — Experiments of Dr. Andrews. — Viscosity. — The Three States of Matter. II. Units of Measurement . . . . . . 8 Units of Length, Area, and Volume. — C.G.S. System of Units. — Units of Mass and Weight, — Derived Units. — Geometrical Relations. Ill Density, — Specific Gravity . . . ^ , 12 Definitions and Measures of Density and Spe- cific Gravity. — Ordinary Definition of Specific Gravity. — Tables of Specific Gravity. —Density and Specific Gravity of Compound Substances. — Exercises. CHAPTER IL FLUID PRESSURE ON SURFACES IMMERSED. IV. Explanation of Terms. — FascaTs Principle . 20 Intensity of Fluid Pressure. — Variation of Pres- sure with Depth. — Direction of Pressure. — Equal Transmissibility of Fluid Pressure. — Mechanical Appliances — Communicating Vessels. — Exer- d by Google X Contents^ SECTION PAGf V. Whole Pressure on Surface Immersed . . .30 Definition and Measure of Whole Pressure. — Pressure on Base and Sides of a Vessel containing Liquid. — Pascal's Vases, — Exercises. VI. Centre of Pressure -37 Definition. — Centre of Pressure of Rectangular Area — Of Triangle.— General Expression. — Examples. — Exercises. CHAPTER III. FLUID PRESSURE ON BODIES IMMERSED. VII. Resultant Vertical Presmte. — Principle of Arcki- medes 44 Determination of Resultant Vertical Pressure. — Principle of Archimedes. — Experiments. — Real and Apparent Weight of Bodies. — Relative Gra- vitation of a Body in a Fluid. — Exercises. VIII. Floating Bodies, — Metacentte . . . . £i Principle of Flotation. — Stable and Unstable Equilibrium. — Definition of Metacentre. — Examples. — Exercises. CHAPTER IV. SPECIFIC GRAVITY, AND MODES OF DETERMINING IT. IX. Application of Principle of Archimedes to the deter- mination of the Specific Gravity of Bodies . 61 Specific Gravity of a Heavy Solid insoluble in Water.— Of a Solid that Floats in Water.— Of a Liquid.— Of a Solid soluble in Water.— Of Gases. — Exercises. d by Google Contents. xl SECTION PAGE X. Other Methods of Determining Specific Gravity.'^ Hydrometers ...... 66 The Specific Gravity Bottle.— Fahrenheit's Hy- drometer.— Nicholson's Hydrometer. — Exer- cises. CHAPTER V. THE MOTION OF LIQUIDS. XI Liquids momng by their own Weight 7 1 Definition of Pressure Height or Head of Liquid. — Torricelli*s Theorem. — Relation of Velocity of Flow to Sectional Area of Vessel. — Vena Contracta. — Liquid Flowing through a Pipe of Variable Area. — Relation of Velocity of Flow to Pressure. — Flow of Liquid through Small Orifice. — Effect of Friction on Pressure at Diffe- rent parts of a Pipe. Xil. Capillarity 86 Drop Formation. — Experiments. —Surface Ten- sion. — Capillary Phenomena, Experiments.-- Principles of Capillarity. — Law of Diameters. XIII. Diffusion of Liquids 94 Graham's Experiments. —Crystalloids and Col- loids. — Dialysis. — Osmose. — Structure of Li- quids. CHAPTER VI. THE PRINCIPLES OF PNEUMATICS. XIV. General Properties of Gases, — Atmospheric Pressure 102 Definition of Pneumatics. — Expansibility and Compressibility of Gases. — Experiments showing that the Air has Weight. — Measure of Atmo- spheric Pressure. — Torricelli's Experiment. — Barometers. — Barometric Corrections. — Abso- lute Pressure per Umt Area. — The Siphon. — Exercises. d by Google xii Contents. SECTION ^^®^' XV. BoylisLaw . . . • • * * "4 Experiments for Pressures greater than Atmo- spheric Pressure.— Results of Expenments.— For Pressures less than Atmospheric Pressure. — Dalton's Law. — Graphic Representation of Boyle's Law.— Limits of Boyle's Law; Expen- ments of Regnault and Despretz.— Relative Densities of the Air at different Heights.— Use of Barometer in Determining Heights.— Exercises. XVI. Diffusion of Gases '^9 Experiments. — Rate of Diffusion. — Graham's I^w.— Kinetic Theory of Gases. CHAPTER Vn. PNEUMATIC INSTRUMENTS. XVn. Diving Bell— Pressure Gauges I'^e 141 Description of and Problems on Divmg Bell. —Manometers.— Compressed Air Manometer. —The Siphon Gauge.— Exercises. XVni. Air-Pumps Parts of a Pump.— Single-Barrelled Air-Pump. —Density of Air after a certain number ot Strokes.-Difficulty of Working. -Double- Barrelled Air- Pump. -Tate's Air-Pump.- Magdeburg Hemispheres. - bprengels Air Pump.-Condensing Syringe. —Exercises. XIX. Pumps for Liquids 5 Common or Suction Pump. — Tension of Piston-Rod.-The Lifting Pump^The For- cing Pump. -With Air Vessel.— Fire-Engme. Bramah's Press.— Exercises. Miscellaneous Problems 162 d by Google HYDROSTATICS. CHAPTER I. PRELIMINARY NOTIONS. I. Nature of Fluid Bodies, § I. Hatter. — What we call matter may exist in the solid form as iron, wood, and ice, or in the fluid form as water, oil, air, and steam. § 2. Essentjial differences between Solid and Fluid Bodies. — If we take any portion of a solid body, such as a piece of metal, a sheet of glass, or a block of wood, one of the first things we observe is that it possesses a definite shape, which cannot be changed except by the application of a certain amount of force ; we may also observe that it occupies, wherever it may be placed, the same amount of space. If, however, we take a given portion of a fluid substance, such as water or air, we find that it possesses no definite shape, and that it moulds itself to the form of the vessel in which it is contained. Thus if we pour a certain quantity of water from one vessel into another, we observe that whilst the volume of the liquid remains Digitized by VjOOQ IC 2 Hydrostatics, the same, its shape changes with that of the portion of the vessel which it occupies. If, again, we endeavour, as in the act of cutting, to separate one part of a solid from another, a certain amount of pressure, depending on the nature of the material, must be exerted ; but if we pass a smooth plane surface, such as the blade of a knife or a sheet of glass, in the direction of its plane, through a mass of fluid, very little resistance is experienced. These experiments show that the particles of a fluid are mobile^ i.e., they move freely among one another, and cohere so feebly that they can be separated from oae another by the application of a very slight force. We may, therefore, define a perfect fluid as a substance the particles of which move freely among one another. If we take a straight cylindrical lube, open at one end and closed at the other, and fit into it a smooth rod of some solid substance, such as iron or wood, and press the free end of the rod in the direction of its length, the pressure applied is transmitted to the closed end of the tube without producing any effect on its curved surface ; but if the tube contain a fluid instead of a solid, and pressure be applied to it by means of a piston fitting into the tube, the pressure will be felt not only at the base of the vessel, but like- wise at all points in its curved surface. This difference, which is characteristic of these two states of matter, is also due to the fact that the particles of a solid, being more or less rigidly connected, are capable of holding together under the influence of a certain amount of force, whilst those of a fluid, being to a Digitized by VjOOQ IC Liquids and Gases. 3 great extent independent of one another, tend to move away in all directions when acted upon by a force in any one direction. For this reason fluid bodies under the action of gravity cannot rest on a hard horizontal surface, otherwise unsupported, as solids do, but re- quire in addition lateral support. Hence the follow- ing definition : FMid bodies are those which cannot sustain a longitudinal pressure, however small, without being supported by lateral pressure also. § 3. Two kinds of Fluids. — There are two great classes of fluids, called liquids and gases, which are thus distinguished : — If a given quantity of water, which is liquid, be poured into a vessel, it will occupy a certain portion of the vessel and no more ; but if a small portion of hydrogen or carbonic acid, which are gases, be intro- duced into a vessel, however large, the gas will expand so that some of it will be found in every part of the vessel. If we take a cylindrical vessel fitted with ^<^ »• a piston and fill it with water, we shall find HT that no amount of pressure we can apply j— 1~ will sensibly diminish the volume of the water; but if, the vessel being filled with air, we press down the piston, the volume occupied by the gas will be found to diminish as the pressure is increased. Liquids, when submitted to very considerable pres- sure, have been found to undergo some diminution of volume, but so little, that for most practical purposes they may be considered as incompressible fluids. Hence a. perfect liquid may be defined as a mass Digitized by VjOOQ IC 4 Hydrostatics. which is absolutely incompressible and absolutely devoid of resistance to change of shape. Matter, however, satisfying this condition does not exist in nature. The chief difference between the liquid and the gaseous state of matter consists in this, that whilst a given portion of a liquid has a definite volume, but no definite shape, a given portion of a gas has neither defi- nite volume nor shape, its volume and shape being always the same as that of the vessel containing it § 4. Change from Oaseons into Liquid State. — Some substances exist at ordinary temperatures both in the liquid and gaseous state. Thus we have steam and water, and ether both as a liquid and as a gas or vapour. Other gases cannot be reduced to the liquid state except under the influence of extreme cold and great pressure. The temperature at which the change takes place, and the amount of pressure required, vary considerably. Until very recently, all efforts had failed to reduce certain gases to the liquid condi- tion ; and consequently these gases were called permanent gases, as distinguished fi-om vapours or liquefiable gases. But the experiments of MM. Pictet and Cailletet, performed in December 1877, have demonstrated as a fact what was previously only an inference firom analogy, that every gas is the vapour of some liquid, and can be reduced to a liquid under the necessary conditions of temperature and pressure. A few days only after M. Pictet of Geneva had suc- ceeded in liquefying oxygen, M. Cailletet of Chatillon- sur-Seine liquefied not only oxygen and carbonic oxide, but likewise hydrogen, nitrogen, and air. Digitized by VjOOQ IC Gaseous and Liquid States. 5 Although there is thus no absolute distinction between permanent gases and vapoiu^, it is convenient to use the word * vapour ' to indicate a gas which at ordinary temperatures can be reduced to the liquid state. Experiments by Dr. Andrews have shown that gases in changing into liquids can be made to pass through an intermediate condition, in which it is im- possible to say to which of these two states of matter they more exactiy correspond. By enclosing a vapour in a tube, and subjecting it, at a very high tempera- ture, to a considerable pressure, the vapour can be made to pass by imperceptible degrees, i.e. without any apparent optical change, into the liquid state. For this purpose the vapour must be compressed to that volume which it would occupy in the liquid state, the temperature being sufficientiy raised to prevent liquefaction from taking place. At a particular tem- perature, which is known as the critical point, and is diflferent for different gases, the tube is found to be occupied by a homogeneous fluid which is neither a liquid nor a gas, but which changes into one state or the other by slightiy lowering or raising the tem- perature, the volume remaining constant We thus see that liquids and gases are convertible the one into the other, and that matter can pass from one state to the other without any perceptible gradations. § 5. Viscosity. — Fluids differ very widely with respect to their distinguishing characteristic, viz. the mobility of their particles. In a perfect fluid the par- ticles are supposed to move among one another with- out encountering any frictional resistance such as Digitized by VjOOQ IC 6 Hydrostatics, retards the motion of one solid when moving on the surface of another. But no fluid exists which fulfils this condition. By agitating a fluid, that is, by causing one part of it to move against another, heat is gener- ated just in the same way as when two sticks of wood are rubbed together. This shows that the particles encounter frictional resistance to their motion. Gases approach more nearly to the definition of a perfect fluid than liquids. The latter are found to exhibit every variety of difference with respect to the mobility of their 4)articles, some approaching to a semi-solid condition, and exhibiting properties inter- mediate between the liquidity of water and the rigidity of ice. Liquids such as treacle, new honey, and tar, in which this frictional resistance appreciably interferes with the mobility of the particles, are called viscous. This viscosity, or * quasi- solidity,' as it is sometimes called, is common to all liquids, and exists, though to a small degree, in water. To this property is mainly due the resistance which a vessel experiences in its passage through the sea ; and it can be shown that a body, completely submerged, and moving with a uniform velocity through a perfect fluid, would experience no resistance whatever to its motion. § 6. The Three States of Matter.— If we compare together certain typical examples of solid, liquid, and gaseous bodies, such as stone, water, and air, we find that they exhibit distinct and characteristic properties, by which they may be referred to different classes. Keeping these differences in view, we have seen that we can define a perfect fluid or a perfect liquid, al- though we know very well that matter nowhere exists Digitized by VjOOQ IC The Three States of Matter, 7 which fulfils the conditions of these definitions. But when we consider the varieties of solid, liquid, and gaseous bodies which experience brings under our notice, we discover that matter actually exists in all intermediate states between these three typical con- ditions. Thus we have glue-like liquids of every kind, between a clear limpid liquid and a gelatinous or semi-solid mass ; and we have vapours in that critical condition through which they can be made to pass by imperceptible degrees from the gaseous to the liquid state. No strict lines of demarcation can therefore be drawn between these various conditions of matter ; and the solid, liquid, and gaseous states may be regarded as only widely separated forms in which, under different conditions, the same substance may exist § 7. Hydrodynamics defined. — The science which treats of the application of the laws of Dynamics to fluid bodies is generally known as Hydrodynamics. The axioms, or fundamental principles, of the science are Newton's laws of motion, which apply equally to all branches of Dynamics. The mobility of the particles of fluid bodies gives rise to important diflerences be- tween their behaviour and that of solid bodies under the action of external forces, which render convenient the separate treatment of this subject. Under the general head of Hydrodynamics are included Hydrostatics and Pneumatics, or the study of the laws of motion in their application to liquid and gaseous bodies respectively. d by Google 8 Hydrostatics, II. Units of Measurement, In the solution of problems in Hydrostatics a knowledge of the weights of certain volumes of fluid is very generally required. It is desirable, therefore, to note at the outset the different standards of measure- ment which are commonly employed. § 8. Units of length. — The imit of length gene- rally used in England is one foot, which is one-third of a yard. The imperial yard is an arbitrary measure- ment, not derived from any fixed quantity in nature, and is defined * as the distance between two marks on a certain metallic bar preserved in the Tower of London, when the whole has a temperature of 60° R' In the French or metric system the stan^rd is the metre, * defined originally as the ten-millionth part of the length of the quadrant of the earth's meridian from the pole to the equator ; but now defined prac- tically by the accurate standard metres laid up in various depositories in Europe.' The metre is some- what longer than the yard, being equal to i '093623 11 yard, or to 39*370432 inches. The great convenience of this system for ordinary purposes is the employ- ment of decimal parts or multiples of the metre, to represent smaller or larger units. Thus, in any ex- pression, if the units represent metres, the tens re- present rt^^f^^-metres, the hundreds hecto-mtixt^, and so on. In the same way the first decimal place represents deci-mttrts, the second cenfi-mtXxtSy the third mi//i-mQtres, and so on. Thus 135724 metres represents i hecto-metre, 3 deca-metres, 5 metres, 7 deci-metres, 2 centi-metres, and 4 milli-metres. Digitized by VjOOQ IC U7tits of Length and Volume. 9 In physical investigations the centimetre is now generally accepted as the unit of length. This unit has been selected by a Committee appointed by the British Association, and is recommended for general adoption in a work * published by the Physical Society of Lon- don. The system of units, based on the recommen- dation of this committee, is known as the Centimetre- Gramme-Second system of units, and is generally referred to as the C.G.S. system. I foot = 30*4797 cm. When the number of units is very large, it is expressed as the product of two factors, one of which is a power of 10. Thus 3240000000 is written 3*24 X 10®, and 0*00000324 is >vritten 3*24 x io~®. § 9. Units of Area. — In England we use, com- monly, the square yard, the square foot, and square inch. In the metric system we have the sq. metre, the sq. decimetre, sq. centimetre, &c. I sq. metre =100 sq. decs. = 10,000 sq. centimetres. In the C.G.S. system of units, the unit of area is the square of the unit or length, i.e. i sq. centimetre. I sq. foot =: 929*01 sq. cm. § 10. TTnits of Volume. — The advantages of the metric system over that ordinarily used in this country are most apparent when we have to compare units of volume with units of length. In our system of mea- sures no simple relation exists between these two units. The gallon, which is the common unit of volume, cannot be represented by any exact number of cubic feet or inches. In the French metric system the unit • * Illustrations of the C.G.S. System of Units,' by J. D. Digitized by VjOO^ \SL lO Hydrostatics. of volume is the litre^ and a litre is equal to a cubic decimetre. Thus : — I cubic metre = looo litres. I gallon = 4*54346 litres = 277*274 cubic inches. In the C.G.S. system the unit of volume is the cube of the unit of length, i.e. the cubic centimetre. I cubic foot = 28316 cubic centimetres. § II. ITnits of Mass and Weight.— The British unit of mass is the quantity of matter which weighs one pound. It is defined by standard only. The French standard is the kilogram, defined originally as the quantity of matter in a litre of water at 4° C, but now practically determined as the mass of a particular piece of platinum preserved in the Ministfere de ITntdrieure at Paris, and by standards which have been compared with this. In the C.G.S. system, the unit of mass is one gram, and is equal to the mass of a unit-volume, i.e. a cubic centimetre of water, at 4° C. I grain = '0647990 grams. As the weights of bodies are proportional to their masses at places equally distant from the earth's centre, the foregoing units of mass may be taken as equivalent units of weight. Thus the weight of 3 cubic centimetres of water is 3 grams, and the volume occupied by 3,000 grams of water is 3,000 cubic centimetres. § 12. Unit of Time ; derived Units.— The unit of time is one second. Units of velocity, acceleration, momentum, force, energy, heat, &a, which are based Digitized by VjOOQ IC Preliminary Notions. 1 1 on the fundamental units of length, mass, and time, are called derived units. Thus in the C.G.S. system the unit of force is that force which, acting upon a gram for one second, generates a velocity of one centimetre per second. § 13. Geometrical Eelations. — ^As problems fre- quently occur which presuppose a knowledge of the measurements of the areas and volumes of certain figures, the following geometrical relations should be remembered : — (i) The ratio of the circumference of a circle to its diameter = 3-14159 = fjg = \^ nearly, and is represented by the Greek letter x. (2) The circumference of a circle = 2 tt r, where r is the radius of circle. (3) The area of a circle = ir r*. (4) The area of the surface of a sphere = 4 ir r*. (5) The volume or contents of a sphere = | tt ^^. (6) The area of the curved surface of a cylinder equals the product of the height into the circumference of the base = 2 tt rh, (7) The volume of a cylinder equals the product of the height into the area of the base, = TT r'^,h, (8) The area of the curved surface of a cone equals the product of the slant side into half the circumference of the base, = tt rs/h^-^r^y where h is the height of the cone. (9) The volume of a cone equals one-third of the volume of a cylinder on the same base, and of the same height, = 3^ t r'^,h. Digitized by VjOOQIC 12 Hydrostatics. III. Density, Specific Gravity, § 14. Density. — If we take two vessels of equal capacity and fill the one loosely with some substance, such as sand, and compress into the other a much larger quantity of the same substancCy we should say that the density of the matter in the one vessel was less than that in the other. If, then, we understand by mass quantity of matter, we see that the densities of two bodies of equal volume and of the same material are proportional to their masses. § 15. Measure of Density. — The measure of density is the mass of a unit- volume. If we adopt the cubic centimetre as the imit- volume, and the gram, or quan- tity of matter in a cubic centimetre of water at 4° C., as the unit of mass, then the density of a body is measured by the number of grams in a cubic centi- metre of its substance ; and if //represent the density of a body whose volume is V and mass M^ d is the M mass of a unit volume, and Jf = Vd\ or // = — . § 16. Specific Gravity. — If we take two vessels of equal capacity, and fill the one with mercury and the other with water, we shall find that the one con- taining the mercury is heavier than the one filled with water. This difference in the properties of the two substances is known as a difference in their specific gravities. When we say that lead is heavier than wood, we mean that bulk for bulk the one substance is heavier than the other — that a cubic foot of lead weighs more than a cubic foot of wood. § 17. Measure of Specific Gravity.— The sp. gr. Digitized by VjOOQ IC Density and Specific Gravity, 13 of a substance is measured by the weight of a unit- vohime of that substance. If d be the mass of a unit- volume, and s its weight, then s=:gd^ where ^ is the acceleration due to gravity ; and if W^ be the weight of a body whose volume is V and sp. ^. s, then W=Vs,oxs=: ~ Since, also, s = gd, we have JV^gd, V, If we adopt the weight of a gram as the unit of weight, the specific gravity of a body is expressed by the weight in grams of a cubic centimetre of its substance. Ordinary definition of Specific Gravity. — The spe- cific gravity of a substance is very often said to be measured by the ratio of the weight of a given volume of that substance to the weight of an equal volume of some standard substance ; and in considering solid and liquid bodies, water at 4° C. is taken as the stan- dard ; whilst in the case of gases, air at 0° C. and 76 cm. barometric pressure is employed. But it will be seen that by defining specific gravity as the weight of a unit- volume, we avoid the explicit reference to a ratio, whilst the number expressing the ratio, when water is the standard substance, is the same as the number of grams representing the specific gravity. Thus, if W be the weight of a given volume V of any sub- stance, and W the weight of the same volume V of water, then according to the ordinary definition W -^ W '=' specific gravity. But if s be the weight of a unit- volume of the substance, and a/ the weight of a unit- volume of water, W^=^ Vs and W^ = Viv. Vs s Hence, specific gravity = -- - = — , and if wc take Vw w Digitized by VjOOQ IC 14 Hydrostatics. the weight of a unit- volume of water to be the unit weight, as we have supposed, then the specific gravity of a substance = j, the weight of a unit-volume of that substance. Since, also, the weight of a unit of mass is equal to g units of force, and is represented in gravitation units by one gram, we see that the numbers representing the densities of bodies represent also their specific gravities, and that the specific gravity of water is unity. When the specific gravity is considered as a ratio, it is sometimes called the relative specific gravity y to dis- tinguish it from the absolute specific gravity ^ or weight of the mass of a unit- volume. Where, as in the English system of weights and measures, the weight of a unit- volume of the standard substance is not adopted as the unit of weight, the specific gravity of any substance, considered as a ratio, = J" -f- a/, where s is the weight of a unit- volume of the substance, and w is the weight of a unit- volume of the standard. If, therefore, S be the relative specific gravity of the substance, we have ^5= Vs-^w or VSw =zVs = lV. Thus, if we wish to find the weight of 4 cubic inches of zinc, we must first know the weight of a cubic foot of water (a/), and then the specific gravity of zinc (represented by a ratio) being 7, we have JF= 7 X ttVb- ^ ^y ^^d taking w to equal 1,000 oz. roughly, we have H^= 7X4xiooo ^ ^.^^ 1^^^ ^^^^ 16x1728 The weight of a cubic foot of water is more nearly equal to 62*3 lbs. In estimating the weights of gases, it is useful to Digitized by VjOOQ IC Tables of Specific Gravity. ^S remember that one cub. cm. of dry air at o° C. and 76 cm. pressure (at Paris) weighs 0*001293 grams. Under the same circumstances, 13 cubic feet of air weigh very nearly i lb. avoirdupois. § 18. SPECIFIC GRAVITY OF SOME IMPORTANT SUBSTANCES. TABLE I. Solids. Name of Substance Specific Gravity Name of Substance Specific Gravity Platinum, cast Gold, cast . Lead, cast . Silver . . . Bismuth . . Copper, hammer^ „ wire Brass . . Nickel . Steel . , Iron, wrought Iron, cast Tin . . Zinc . . Antimony Iodine Diamond . Flint-glass Aluminium Bottle-glass Plate-glass Marble . Emerald . Rock-crystal Porcelain 20-86 1925 11-35 10-47 9-82 8-88 878 8*39 8-28 7-82 779 7*21 7-29 7 'CO 671 4*95 3-52 378 to 3-2 2-57 2-6o 2-37 2-84 277 2-66 2*49 to 2*14 Sulphur, native Ivory .... Graphite . . . Phosphorus . . Magnesium . . Amber . . . Wax, white . . Sodium , . . Potassium . . Ebony, American Oak, English . Mahogany, Spanish Box, French Beech . . Ash . . Maple Cherry-tree Walnut . Pitch pine Elm . . Cedar Willow . Larch . . Poplar Cork . . 203 1*92 1*8 to 2*4 177 174 I -08 097 0-97 0-86 1*33 o'97 to I '17 I '06 1*03 0-85 0-84 075 071 0-68 0-66 060 059 0-58 0-54 038 0*24 d by Google i6 Hydrostatics. TABLE II. Liquids, at o® C. Name of Substance Specific Gravity Name of Substance Specific Gravity Mercury .... Sulphuric acid . . Nitric acid . . . Aqua regia . . . Hydrochloric acid . Blood, human . . Ale, average . . Milk Sea-water . . . Vinegar .... Tar 1-848 1*500 1*234 I -218 1*045 1*035 1-030 I '028 1-026 I 015 Water, distilled, at Linseed oil . . . Proof spirit . . . OHve oil ... . Ether, hydrochloric Turpentine, oil of . Brandy .... Alcohol, absolute . Ether, sulphuric . I 000 0-940 0-930 0-915 0-874 0870 0-837 0796 0720 TABLE III. Gases at 0° C. and 76 cm. Pressure. Name of Substance Specific Gravity Name of Substance Specific Gravity Oxygen .... Atmospheric air Nitrogen .... Hydrogen . . . Chlorine .... 0001432 0001293 0001267 0000894 OX)03209 Hydrochloric acid gas Nitrous oxide . . Carbonic acid . . 0-00164 0-00197 0-00198 d by Google Density and Specific Gravity of Compounds, 17 § 19. To find the density of a combination of two or more substances whose volnmes and densities are ^ven. Let z/j, z/2, v^ ... be the volumes of the substances, d\^ ^2> ^3«- their respective densities. Then if Fbe the volume of the combination, and if no contraction take place, V=Vi +V2+ Vq 4- ... and if Z) be the density of the whole, V£> equals the mass of the whole, and therefore, VD = Vi di +^^2 ^2 + ^'3 ^3 + ••• or Z) = i^lAjt" ^2 ^2 + z^3 ^8 + '" Vi + >2 + v^ +... If, however, as very frequently happens, contraction takes place, and if the volume of the whole is some proper fraction (= r) of the sum of the volumes of the parts, v,d, ^ v^d^^v^d^ + ... ^^^■^-r(z/i + z;, + z/3...) A similar proposition holds good if, for density, we substitute specific gravity. § 20. To find the specific gravity of a combina- tion of substances, the weights and specific gravities of which are given. Let ze/i, 0/3, a/3 ... be the weights of the components, and ^1, J2, s^ ... their respective specific gravities ; then, if W be the weight of the whole and S its specific gravity, we have IV = w^ + 0/3 + 0/3 4- ... G Digitized by VjOOQ IC iS Hydrostatics. and since W = VS^ where V is the volume of the whole i^ — . ^ -4- -^ 4- J£a 4- O Jj ^2 "^3 supposing no contraction to take place, and there- fore «$"! '$'2 "^3 But if the volume of the whole be less than the siim of the volumes of the parts in the ratio of r : i, then 2— ^\ 4- «^2 + ^3 + ... ^\S^ ^2 ^3 ^ The density of a compound can be found in the same way, the masses and densities of the components being given. § 21. Examples.— (I.) i,ooo cc. of a gas whose density is 12 are mixed witb 2,cxDO cc. of a gas whose density is i6, and the volume of the mixture is diminished by one-third. Find the density of the mixture. T^ 1000x12 + 2000x16 U B _ = 22 f(l000 + 2000) (2.) Three kilograms of a substance, sp. gr. = 4, are melted with 5 kils. of a substance, sp. gr. = 6, and the volume of the mixture is o*i less than the sum of the volumes of its com- ponents. Find sp. gr. of mixture. ^ = 3 + 5 _ 8x240 _ -35 &(* + g) "" 9x38 '^^ Digitized by VjOOQ IC specific Gravity of Compounds, 19 Exercises. I, 1. Find the weight of a bar of copper, the section of which is 64 square centimetres, and length one metre. ^ 2. Find the size of an iron shot that weighs 60 lbs. ^/ v, c^c - 3. Find the weight of a column of mercury 30 inches high, having a uniform section of one square inch. 1 1^ r* 4. A cubic inch of a substance weighs 2 oz. ; find its specific gravity. J. ^ 5. If silver and copper are mixed in the proportion of 2 : 7 by weight, find the specific gravity of the compound f y Q 6. What mass of copper must be mixed with 200 grams of gold to make the specific gravity of the compound 0*9 of that of gold? >.:^,^^W.-^ 2,/. /)3*w- 7. Equal weights of two fluids, of which the specific gravities are s and 2j, are mixed together, and the mixture occupies three- fourths of the sum of the volumes of its components. Find the specific gravity of the mixture. ^)^ 8. Three litres of a gas, the density of which is unity, are mixed with one litre of a gas density 14, and the mixture occu- pies half the volume of the original gases. . Find its density. ?i-^ 9. A bar of cast-iron, sp. gr. = 7*2, is found to weigh 17-5 kils. The length of the bar is 5 decimetres, and sectional area 50 sq. centimetres. Is there a flaw in the casting ? If so, what Is its size? l^Qj^ ^^^ C3 Digitized by VjOOQIC 20 Hydrostatics. CHAPTER 11. FLUID-PRESSURE ON SURFACES IMMERSED. IV. Explanation of Terms. PascoTs Principle, § 22. Fluid Pressure. — The base and sides of a vessel containing a liquid in equilibrium are subjected to a certain pressure, which is caused by the weight of the liquid, and by the resistance which the surface of the vessel offers to the free motion of the liquid. § 23. Intensity of Pressnre.—By intensity of pressure at a point is meant the pressure on the unit of area containing that point If a small surface a^ in contact with a fluid, is maintained in equilibrium by a force jP, acting perpen- p dicularly to the surface, then - measures the average a pressure-intensity at any point of the area ; and if P -= /, then/ is the pressure on a unit of area. a § 24. Variation of Pressure with DeptL— Sup- pose the small horizontal area a to be at a depth i below the surface of the liquid, then the area a will be pressed vertically downwards by a force equal to the weight of the column of liquid above it ; and if x equal the weight of a unit volume of liquid, this force is equal Digitized by VjOOQ IC Variation of Pressure with Depth, 21 io z as. If, then, / be the pressure at any point of this area, p .a^= zas, orp = z s. If a is not horizontal, the pressure at some parts of the area is greater than at others. But if we sup- pose this area to become smaller and smaller, and ultimately to diminish without limit, the difference of pressure at different parts of the area will diminish likewise ; and if / represent the pressure intensity, when the area has become so small that it may be regarded as a point, then/. a^=z.a. s, where a is the infinitely small area about this point, and z is its depth. Hence/ = z,s, as before, showing that /;/ t/ie same liquid t/ie pressure varies with the depth. This proposition may be experimentally verified by taking a small cylindrical vessel a (fig. 2), open at both ends, and having a movable base o, to which a thread c is attached. If we connect the base with one end of a scale-beam and press y\c, a. it against the vessel by at- taching a weight W to the other end of the balance, we shall find, on pouring water into the vessel, that the disc falls off when the water has risen to a certain height, and that this height is always pro- portional to the excess of the weight W above the weight ' of the disc. This shows that the ratio of the pressure on the disc to its depth below the surface of the liquid is constant, />. that the pres- sure varies with the depth. Digitized by VjOOQ IC 22 Hydrostatics, By fixing the vessel a into another vessel contain- ing water, it may be shown that the liquid in the outer vessel exerts an upward pressure on the under side of the disc equal to the downward pressure of the liquid contained in the inner vessel. For, if we attach the disc o to one end of the scale-beam, and to the other end a weight sufficient to counterbalance it, we shall find, on pouring water into a, that the disc does not fall off until the level of the water in the two vessels is the same, § 25. Direction of Pressure on a Surface in Con- tact. — If a thin plate be immersed in a fluid and be held in any position, the direction ot the pressure is perpendicular to the surface of the plate, if the fluid is at rest. For if it acted in any other direction it could be resolved into two components, one per- pendicular to the surface and the other along it; and the latter component would, in the absence of friction between the surface and the fluid, produce motion, which is contrary to supposition. The direction of the pressure must, therefore, be perpendicular to the surface. § 26. Equal Transmissibility of Fluid Pressure. If we take a vessel full of water (fig. 3), having various apertures of the same size, fitted with water- tight pistons which are kept in equilibrium, and if one of these a be pressed downwards with a force p, an additional pressure equal to p will be required at each of the other pistons to preserve equilibrium. Thus if a force of i lb. be applied at a, it will be necessary to apply an increased pressure of i lb. to each of the other pistons to prevent motion ; and as Digitized by VjOOQ IC Equal Transmissibility of Fluid-pressure, ^3 it matters not in what part of the vessel these pistons Fig. are fitted, we see that any increase of pressure applied at one part is transmitted equally throughout the fluid. If the pistons, as in fig. 4, are of different areas, that of B being twice that of a and the area of c being nine times as great, it will require an additional force of 2 lbs. at B, and of 9 lbs. at c, to preserve equilibrium, when a force of i lb. IS applied at a. These experiments serve to illustrate the following fundamental law of fluid-pressure, which is known as Pascal's ^ principle : When pressure is communicated to any part of a fluids it is trans- mitted equally in all directions through the fluid. § 27. Hechanical Application. principle, which is a ne- necessary consequence of the mobility of the par- ticles of a fluid, serves to explain the action of a very useful mechanical contrivance for multiply- ing power. * Pascal was bom at Auvergne, 1623 ; dkd 1662. Digitized by VjOOQ IC 24 Hydrostatics. It consists of two communicating vessels contain- ing water, one of which is much larger than the other. The vessels are fitted with pistons P and /, the areas of which we will suppose to be A and a. If now weights ?f^ and zc' be placed on those two pistons respectively, so as to counterbalance one another, it will be found that W \w\\ A : a, which is in accordance with Pascal's principle. We see, also, that if the piston / be pressed down through the space S, the water contained in the smaller vessel will pass into the larger, and force up the piston F through some space s, such that— ax SssAxs, since the volume of water that is removed from one vessel is the same as that which enters the other vessel. Hence S^A_W saw or, w, S=i W, s, />., the work done by 7V = work done by W. § 28. Hydrostatic Paradox. — A consequence of Pascal's principle is that a quantity of water, however small, can be made to support a weight, however large, and this seeming paradox can be exhibited in the following manner. Let A B (fig. 6) be a long narrow pipe communi- cating with a vessel c d, into which a piston c e is fitted. If now water be poured through the pipe, it will be found to rise to the same level, c e f, in both parts of the vessel ; but if a weight Wht placed on c E, and additional water be poured into the pipe, Digitized by VjOOQ IC Flow dependent on difference of Surf ace4eveL 25 a position of equilibrium will result such as is shown in the figure. In this case, the pressure exerted by the water in the pipe above the plane c e f supports the weight W. Since this pressure is communicated to every element of area of c e equal to the sectional area of the pipe, the whole pressure on c E is as many times greater than the weight of the water in A F as the area of c e is greater than the area of the pipe. Let a equal the sectional area of the pipe ; then if P is the pressure at f produced by the weight of water above it, /^ = dJ x a f x a/, where w is tl\e weight of unit-volume of water, and if A is area of piston c E, the pressure communicated to c e is — XaXAF XW= AXAFXW a /. W =iA X A F X a/. This shows that the weight IV can be made as great as we please, by taking A f or c e sufficiently great, and that it is independent of «, the section of the tube. Hence, with a tube sufficiently narrow, a quantity of liquid, however small, can, in principle, support a weight, however large. § 29. Communicating Vessels containing Liquid. If two vessels a and b (fig. 7) containing the same liquid be connected by a pipe, and if the surface level Digitized by VjOOQ IC 26 Hydrostatics, of the liquid in a be higher than that in B, the liquid will be found to flow from a to b. This is easily explained by considering the pres- sure on either side of any sec- tional area, / ^, of the pipe. For it is evident that the pres- sure on the side towards A is greater than on the side to- wards B, since the pressure varies with the depth below the free surface, and consequently the flow takes place from A to B. It should be observed that although in the case now considered the liquid flows from a to b, the pressure-intensity along the base of a is less than that along the base of b , and also that the pressure at a^ where the fluid escapes from the one vessel, is less than that at ^, where it enters the other vessel. If we connect by a tube c d two parts of the same vessel, although the pressure at d is greater than that at c, no flow takes place, for it will be seen that the pressure on either side of any element of area in the pipe c d is the same. We see, therefore, that a liquid always flows from places of higher to places of lower surface level, and that if a liquid is in equilibrium its surface level must be uniform. Hence it follows that if a liquid be con- tained in a vessel, its surface must be horizontal ; or, more generally, every point of the surface of a liquid in equilibrium must be at the same distance from the earth's centre. In the case of large inland seas, the surface is somewhat curved. Digitized by VjOOQ IC Communicating Vessels, t.'j We also see that the pressure at all points in the same horizontal plane within a liquid at rest must be the same, since these points are all equally distant from the free surface, and the pressure varies with the depth (§ 24). It also follows that if any number of vessels con- taining liquid communicate with one another, the liquid will stand at the same level in all. For if the surface level were higher in any one vessel, a flow of liquid would take place from that vessel into the others, till the uniformity of surface-level had been established.^ This fact may be experimentally verified by pour- ing water into one of the parts fig. 8. of a vessel similar to that ^^^ M, £/ £f shown in the figure, when it ^^ ^H S M will be found that the free sur- |^^^^^^^^gi faces of the liquid in all parts ^ of the vessel lie in the same horizontal plane. The tendency of liquids to find their own level, />. to flow from places of higher to places of lower surface-level, is of great practical importance. It is utilised in the water-supply of towns. A reser- voir of water is kept at a considerable elevation, and from it pipes proceed in all directions conveying water to any heights below the level in which it stands in the main reservoir. The water-level is an instru- ment which acts on the same principle. It consists of a tube bent at right angles, and furnished at its ^ It should be observed that difference of surface-level cor- responds with difference of temperature in heat, and with difference of potential in electricity. Digitized by VjOOQ IC 28 Hydrostatics. extremities with two glass vessels, and the whole is partly filled with water. Since the liquid has the same surface-level in both branches, any points which the observer's eye detects to be on the same level with the surface of the water in both branches, will be in the same horizontal plane. When a liquid falls from a higher to a lower level, there is a change of potential into kinetic energy, and a corresponding amount of work can be effected This fact is economically employed in water mills. § 30. Examples. — (i.) Find the whole pressure exerted on a horizontal area of 9 sq. cms. which is sunk 125 cms. below the surface of water. The pressure equals the weight of water supported, = 125x9=1125 grams. (2.) Find the pressure- intensity due to a column of 25 cms. of mercury upon which rests a column of 50 cms. of water. The pressure-intensity is the pressure per unit area, i.e, per sq. cm. : = 25 X 13 '6 + 50 = 390 grams = 390 g units of force, the specific gravity of mercury being 13*6. (3.) Find the vertical force necessary to support the hori- zontal base of a vessel containing mercury, if the area of the base is one square decim., and its depth below the surface of the mercury 5 centimetres, neglecting the weight of the base. °=^ d! j« 5 X 100 X 13*6 grams. .•. P^ 6800 grams. (4.) What must be the height of a column of mercury to exert a pressure of 1220 grams per sq. centim. ? d by Google Fluid-pressure on Surfaces immersed, 29 If A be the height required, I220s^ x 13*6 grams. . , 1220 o_^ . , h = — - = 897 cm 136 § 31. Equilibrium of two different Liquids in Conmnmicating Tubes. — If two liquids that do not mix meet in a bent tube, or in two tubes communicat- ing with each other, the heights of their free surfaces above their common surface are inversely proportional to their specific gravities. Let A a' be a plane drawn through the common surface of the two liquids in one of the tubes. Let s and ^ be their respective specific gravities. Let ff and c be the free surfaces of the liquids. Then p^^ if the liquids are in equilibrium, the pressures at a and a' must be the same. The pressure-intensity at a is « ^ x jt, where ab \% the vertical height of b above a. The pressure-intensity at a' is « ^ x j' ^vhere acvst vertical height of c above a'. Hence ab y. s ^ ac x s' or ab : ac W s' : s. Exercises. II. 1. Two communicating vessels contain fluid, and are fitted with pbtons, the diameters of which are 2 inches and 8 inphes respectively. If a weight of 3lbs. is placed on the smaller piston, what weight must be placed on the larger to preserve equilibrium ? ' ' 2. A narrow vertical pipe is attached to a vessel (fig. 6), Which is fitted with a piston the area of which is 2 square deci- d by Google 30 Hydrostatics, metres. If the vessel and pipe contain water, find the height of the water in the pipe when a weight of 40 kils. is placed on the piston, '^ \ ^ 3. A cylindrical vessel contains mercury to the height of 2 inches above the base, and a layer of 8 inches of water resting on the mercury. Find the pressure at any point in the base, taking sp. gr. of mercury to be 13 '6. 4. At what depth below the surface of a lake is the pressure intensity 5 times as great as at a depth of 10 feet, supposing the atmospheric pressure to be equal to the weight of a column of water 34 feet high ? 5. Two liquids that do not mix are contained in a bent lube ; the difference of their levels is 3 ins., and the height of the densenabove their conmion surface is 5 ins. ; compare their specific gravities. 6. If, in the above, the internal section of the tube is one square inch, and the lighter liquid is water ; find the weight of water contained in the tube. V. Whole Pressure on Surface immersed, § 32. Whole Pressure. — When a vessel contains a fluid, or when a body is immersed in a fluid, the fluid exerts a normal pressure at each point of thesur- FlG. 10. face in contact with it. The sum of all these pressures is called the whole pressure on the surface immersed. It should be observed that these pressures act in (Jifferent directions, the pressure at each point being perpendicular to the surface at that point The whole d by Google WItole pressure on a Surface immersed. 31 pressure is the sum of all these pressures, and repre- sents the total strain to which the vessel containing the fluid, or the body immersed, is exposed. § 33. To find the Whole Pressure which a Liquid exerts on a Surface immersed. — Let a b c d be any area in contact with a fig. 11. homogeneous liquid, the free surface of which is the horizon- tal plane abed. Let ef be any element of area so small that the pressure-intensity at all points may be supposed to be constant, and let the depth of this area below the surface of the liquid be z. Then if s be the weight of a unit- volume of the liquid, the pressure exerted one f \% a z s^ where a is the area of ^/ Now the whole pressure is equal to the sum of the normal pressures on all the elements that make up the whole area. Therefore the whole pressure is equal to ai Zi s + a2 Z2 s -^ a^ z^ s + . . . = {^1 Zi -\- a^ Z2 + a^ z^ -\r . . .) s. But, by the properties of the centre of gravity, a^ Zi + ^2 ^2 + ^3 ^3 + • • • = (^1 + ^2 + ^3 + • • •) ^ where z is the depth of the centre of gravity of the whole area below the plane surface of the liquid ; . '. whole pressure ^= A z s, where A is the whole area in contact, z the depth of its centre of gravity below the surface of the fluid, and s the weight of a unitvolume of the liquid. Or, the d by Google 32 Hydrostatics. pressures Azdg where// is density of fluid and g the acceleration due to gravity. Hence, The whole pres- sure on any area immersed equals the weight of a column of liquid which has that area for base, and the depth of its centre of gravity below the surface of the liquid for height. This proposition holds good whatever may be the shape of the vessel containing the liquid ; and what- ever may be the position of the area immersed. In some cases the results to which it leads seem at first sight paradoxical, but they will be shown to agree with the general proposition which we have now esta- blished. § 34. Examples. — (i.) Compare the pressures on the base aiid side of a cube filled with liquid. Let the edge of the cube be a, then area of the base is a\ and depth of centre of gravity below surface is £Z, /. whole pressure on base is a* x a x s^c^s. The area of a side is also a^ ; but depth of its centre of gravity is ~, /, pressure on side istf^x — x j«— j, 2 22 ,'. whole pressure on base =» twice the whole pressure of one of the sides. The pressure on the base acts vertically, and the pressure on the sides horizontally. (2.) An oblong, the edges of which are 6 metres and 8 metres, is immersed vertically with its shorter edge horizontal, and 2 metres below the surface of water ; find the whole pressure on either side. Area immersed is 6 x 8 = 48 square metres. Depth of centre of gravity is 4 + 2 = 6 metres. /, whole pressure is 48 x 6 x j, where s is the Weight of a cubic metre of water, i,e, 1,000 kilograms. /. Pressure - 288,000 kils. Digitized by VjOOQ IC Fluid-pressure on Surfaces immersed. 33 Fig. (3.) A right pyramid, the height of which is 8 decs., has a square base, each edge of which is 5 decs. Required the whole pressure on the base when the pyramid is full of liquid. The area of the base is 25 sq. decs., and the depth of its centre of gravity below the sur&ce of the liquid is 8 decs. Hence the whole pressure on the base is 80 X250OXJS20OX kils., where s is the sp. gr. of the liquid. It is to be observed that the pressure on the base, in this case, is greater than the weight of the liquid the vessel contains, being equal to the weight of liquid in the prism hav- ing the same base, ue, to a column of liquid the height of which is E F, and base A B c D. On the other hand, the pressure transmitted to the stand on which the vessel rests is equal only to the weight of the liquid contained in the vessel ; and consequently the pressure on the base of the vessel is greater than the pressure communicated to the stand. This result which follows directly from the general proposition, requires further explanation, and will be considered in the fol- io viring paragraph. § 35. To find the whole pressure exerted by a liquid on the base and sides of the vessel con- taining it. — ^We have three cases to consider, ac- cording as the sides of the vessel are vertical or slant from the base outwards or inwards. We shall suppose the base of the vessel to be horizontal If the sides are vertical, as in fig. 13, the pressure on the base is evidently equal to the weight of the fluid con- tained in the vessel, and the horizontal pressures are equal Suppose now that the sides are inclined, outwards in fig. 14 and inwards in fig. 15. The pressure at any point o in the side a b is a Digitized by VjOOQIC, .^4 Hydrostatics, force p acting perpendicularly to a b. As this force is prevented by- the resistance of the slant side from Fia 15. E -.^-U r r -^* ^--j causing motion, it produces a reaction equal in mag- nitude but opposite in direction. This reaction / can be resolved into two components, x and y acting at o, X horizontally in both figs., and y vertically up- wards in fig. 14 and I vertically downwards in fig. 15. It thus appears that in fig. 14 part of the weight of the liquid is supported by the sum of the forces y acting at all points in the slant side, and the remainder of the weight of the liquid presses on the base B c. Hence, in this case the pressure on the base is /ess than the weight of the contained liquid. But in fig. 15 the force y acts downwards, and consequently increases the pressure on the base caused by the weight of the liquid ; and hence the pressure on the base of the vessel is greater than the weight of the contained liquid by the sum of the vertical components of the reactions caused by the pressure of the liquid against all points in the slant sides. Again, since the force /, due to the pressure of the liquid, and the components, x and y of the reac- tion of the surface a b, are in equilibrium, they can d by Google Pascal's Vases. 35 be represented by the sides of the triangle a e b. In the same way the forces/', ^',y, acting at q can be represented by the sides of the triangle d f c Hence/ :^:;c::ab:ae:eb and Z' : J'' : ^ : : c D : D F : F c But E B is equal to f c, when the base is horizontal .-. X = x' \ and in the same way the horizontal pressures at all other points can be proved to be equal. Hence the horizontal components of the pres- sure of the liquid against the slant side are equal If the base is not horizontal, then the whole pres- sure on the base can be resolved into vertical and horizontal components, and the algebraic sum of the horizontal pressures on the base and sides of the vessel will still be found to equal zero. § 36. Experimental verification. Pascal's Vases. — In order to verify the general proposition, which we have now proved, viz., that the pressure on the base of a vessel containing liquid is independent of the shape of the vessel, and varies only with the area of the base and the depth of its centre of gravity below the sur- face of the liquid, Pascal contrived an experiment very similar to the following : Take three vessels . p, q, m, of different shapes, having the same circular aperture at their base, and capable of being screwed into the ring of the stand A B. One of the vessels being fastened to the ring, a circular disc d hanging from one arm of the balance is pressed against it by weights placed in the scale-pan hung to the other arm. Water is then poured into the vessel, and an index / marks the D 2 d by Google 36 Hydrostatics. level at which the water stands when the disc falls off If the experiment is now tried with the other Fig. x6. vessels, it is found that the disc falls off when the water rises to the same level. This shows that the pressure on the base of each vessel is the same when the water is at the same height above it Exercises. III. 1. Find the whole pressure on a rectangular surface 6 feet by 4 feet, immersed vertically in water with the shorter side parallel to, and 2 feet below the surface. 2. Find the whole pressure on the curved surface of a vertical cylinder which is filled with a liquid, sp. gr. = I '5, the height of the cylinder being 2 decims., and the radius of the base 7 cms. 3. Show that if a sphere or a cube be filled with liquid the total strain to which it is subjected is three times the weight of the liquid it contains. 4. A flood-gate is 6 feet wide and 12 feet deep. What is the total pressure on the flood-gate when the water is level witir the top? 5. A globe, tha radius of which is 3*5 cms., rests at tho Digitized by VjOOQ IC Centre of Pressure. 3 7 bottom of a vessel 30 cms. in height, foil of water. Find the total pressure on the globe. 6. A conical vessel 2 decimetres high, has a movable base, of 25 sq. cms. area, formed by a disc 2 cms. thick, and the specific gravity of the material being 3 '2, find the force neces- sary to uphold the disc when the vessel is full of water. 7. A smooth vertical cylinder, 2 feet high and i foot in diameter, is filled with water, and closed by a piston weighing 3 lbs. Find the total pressure on the curved surface. VI. Centre of Pressure. § 37. Definition. — When a plane surface is im- mersed in a fluid, the pressures at different points of the surface are perpendicular to it (§ 25), and consti- tute a system of parallel forces of which the whole pressure is the resultant. The magnitude of this re- sultant pressure we have seen how to find, but the point at which it *acts we have not yet determined. This point is called the centre of pressure^ and may be defined as the point of action of the single force equiva- lent to the whole pressure exerted by a fluid on any plane surface with which it is in contccct If a plane surface is immersed horizontally the centre of pressure corresponds with the centre of gravity, but not so if it be immersed in any other position. For in this case the pressures on equal elements of area are not equal, since the pressure varies with the depth (§ 24), and the different elements of area into which the whole surface may be divided are at different depths below the surface of the fluid. Consequently, the centre of pressure, which is the point in, the plane surface at which the resultant of d by Google 3» Hydrostatics. all these forces acts, does not correspond with the centre of gravity of the surface. The term centre of pressure is used with respect to plane surfaces only, since it is not always possible to find a single force equivalent to the resultant action of a fluid on a curved surface. § 38. To find the centre of pressure of a rectan- gpilar area immersed vertically. Let A B c D be a rectangle having its upper side A B in the surface of the liquid Then ifEF be drawn to bisect A B and d c, the pressure will be equally dis- tributed on each side of e f, and the centre of pressure will lie somewhere in this line. To •determine where, take M N bisected by f to represent the pressure at r. Join em, en. Then if we take any point p in e f, and draw q p r parallel to M F N, Q R will represent the pressure at p. For, since the pressure varies with the depth, the pressure at p : the pressure at f : : e p : e f and ep:ef::qr:mn /. the pressure at p : the pressure at f : ; q r : m n and /. Q r represents the pressure at p. Now the problem of finding the point of action of the resultant of a number of forces represented by such lines as q r, acting at all points of e f, is the same as that of finding the centre of gravity of the triangle e m n. But the centre o^ gravity of this triangle is kno\ni to be I> M d by Google Centre of Pressure, 39 at a point g in e f, such that f g = J f e, and therefore the centre of pressure is at the same point, i,e. at a distance of one-third up the middle line from the base. § 39. To find the centre of pressure of a triangle having one of its sides in the surface of the liquid. If the triangle be divided into a number of nar- row horizontal stnps, the pressure on each strip acts at its middle point, and therefore the whole pressure on the triangle acts somewhere in the median line drawn through the middle point of the side in the surface of the liquid. But the pressure on each strip is proportional to its area multiplied by its depth, and the value of this product is constant for every pair of strips at equal distances from the middle point of the median line. Hence the resultant pressure acts at the middle point of this line, or the depth of the centre of pressure is half that of the apex immersed. § 40. To find a general expression for the centre of pressure of a plane surface. Suppose the surface divided by horizontal lines into any number of small elements, then if a be the area of one of these elements and z its depth (which may be considered the same as the depth of its centre of gravity) below the surface of a liquid of sp. gr. J, azs is the pressure on that element of area. Hence, if -s^i, 2^2, . . . . be the depths of the several elements, the whole pressure equals a\Z\S-\'aiiZi^s-\-a^z^sAr Digitized by VjOOQ IC 40 Hydrostatics. and if Z be the depth of the centre of pressure, then by the principle of parallel forces or Z= ^i^i^+^2^^+g3V+. . . . . The general application of this method, as of the similar method for finding the centre of gravity of a surface, requires the use of the integral calculus, but special cases may be determined by ordinary alge- braical processes. § 41. Bzamples. — (i.) To apply the method of § 40 to find the centre of pressure of an isosceles triangle with its apex in the surface of the liquid and its base horizontal. Suppose the triangle to be divided by horizontal lines into narrow horizontal strips. Let h — height of triangle, h its base ; and let s be the sp. gr. of the liquid. Let h be divided into n parts, then the breadth of each of these strips will be -. Also, since F E ; E A : : B D : DA /. F E = — A E, and the area 2h of each strip may be represented by - • A E •— - = — A E, and 2h ft 2n therefore the pressure on each strip may be taken as equal to - / A E y. ^, supposing the depth of the strip below A to correspond with that of its centre of gravity. Now since — , we have n n n n A E has the several values — , ?-» 3 f . d by Google Fluid-pressure 09t Surfaces immersed. 41 z - ± 2n n h_ n h_ 4 h_ 4 I^ + 2» + 3^+ I«+ 22+32 + . 4 236 236 2« 3 6»* (by summation of series.) and if n be infinite each of the quantities — and \ has for its limit zero, ft fr Hence Z ^ ^ , h. 4 Or, the centre of pressure is on the median line, and at a depth of three - quarters of the height of the triangle, from its apex. (2.) Required the mag- nitude and position of the resultant pressure on a flood- gate, the level of the water being different on either side. Let A B be a section of the flood-gate, and let the height of the water on one side be a, and on the other side ^, and suppose h greater than a. Digitized by VjOOQ IC Fia 19. ^ ■^SZ— rzz-:^. e;:::^ ^. IT" z:r:zr:irz:=~- Birz: 7^:zz H ?-s^^ ^ :^ -J j"^ /^2 Hydrostatics, Then, if >6 be the width of the gate, the total pressure on one side will be — 2 2 ^p. and on the other side — ^^5= Q ; and P acts at a point c such 2 that B c=s one-third of Bw, and Q acts at a point D such that B D =: one- third B« (§ 38). The resultant of these forces Q-P equals (^— a') — , and the point E where it acts can be deter- mined by the principle of parallel forces. Thus : * ' 2 2 2 2\ 3 3/ • B ^-t±hist Exercises. IV. 1. A cubical block each edge of which is 5 cm. is sunk in water, with two opposite faces horizontal ; find the difference in the pressures on its lower and upper surfeces. 2. A cylinder 10 inches high contains liquid to the height of 8 inches : find the line of action of the total pressure on the interior surface of the cylinder. 3. A hollow cube is three-fourths filled with water. One of the sides of the cube moves freely about a hinge at the base. Required the force that must be applied at the upper edge of the moveable side and perpendicular to it to keep it in equilibrium. 4. Find the height of a cylinder the diameter of which is 2 feet, so that the whole pressure on the curved surface may be four times as great as the pressure on the base, when the cylinder is filled with liquid. 5. A rectangle is immersed in water with one side in the surface. Show how to divide it by a horizontal line into two parts on each of which the whole pressure shall be the same. Digitized by VjOOQ IC Fluid-pressure on Surfaces immersed, 43 6. A rectangular v«sel has a partition 10 inches high and 8 inches broad. On one side of the partition is water to the height of 4 inches : on the other side alcohol (sp. gr. o*8) to the height of 6 inches. Find the magnitude and position of the resultant pressure on the partition. 7. An equilateral triangular lamina is immersed in water with one side in the surface and the opposite angle 2 deci- metres below its surface. If each side of the triangle mea- sures 6 decimetres, find the total pressure which the water exerts on it. 8. A rectangle a BCD is immersed in water with the side A B in the surface. Find the pressure on each of the tri- angles formed by the diagonal a c, and show where it acts. Show also that the resultant of these two pressures coincides with the pressure on the whole rectangle. 9. Find the height to which water may rise on one side of a wall 2\ metres high and half a metre thick without overthrow- ing it, the specific gravity of the material of the wall being 2. 10. A cylinder, height h inches and diameter 2r, contains three liquids of specific gravities .S",, S^ .S",, which do not mix in layers of equal thickness : find the whole pressure on the base. d by Google 44 Hydrostatics. CHAPTER HI. FLUID PRESSURE ON BODIES IMMERSED. VII. Resultant Vertical Pressure, — Principle of Archimedes, § 42. We have seen (§ 32) that if a body is wholly immersed in a fluid, every point of its surface is subjected to a pressure perpendicular to the surface at that point. Now, all these pressures, acting as they do in various directions, can be resolved into horizontal and vertical components ; and since the horizontal pres- sures equilibrate each other, the resultant pressure must be vertical, and act upwards or downwards. Moreover, since the pressure varies with the depth, it is clear, speaking roughly, that the whole pressure on the lower half of a body is greater than that on the upper half; and, hence, the resultant of all the pressures on a body immersed is a force acting verti- cally upwards. This force is called the resultant vertical pressure, § 43. Experiment. — Take an ordinary weight of one pound, and having attached it to a piece of strong thread, let it hang wholly immersed in water from one of the scale-pans of the hydrostatic balance, as shown in fig. 22. It will now be found to weigh less than Digitized by VjOOQ IC Fluid-pressure on Bodies immersed, 45 I lb., and this difference of weight is due to the re- sultant vertical pressure acting upwards. § 44. Measure of the Besultant Vertical Pres- sure. — If a symmetrical body with vertical sides be immersed in a fluid, we can easily see what the mag- nitude of this resultant pressure is. For, since the downward pressure exerted by the fluid on the body is equal to the weight of the column of fluid having a b (fig. 20) for a base, whilst the upward pressure is equal to the weight of the column having c d for a base, the resultant vertical pressure must be equal to the dif- FiG. 20. Fig. 21. ference between the weight of these two columns, i,e, to the weight of a column of fluid equal to a b c d, i,e, to the weight of the fluid displaced. If the body be of irregular shape, a more general method of proof must be employed. Thus : Suppose the body immersed to be a portion of the fluid itself solidified. Then, if no change of density take place, the solidified fluid will remain as before in equilibrium. Hence the weight of this portion of the fluid, which acts, at its centre of gravity, vertically downwards, must be counterbalanced by the upward pressure of the fluid ; and consequently the resultant Digitized by VjOOQ IC 46 Hydrostatics. vertical pressure equals the weight of the solidified fluid. But the fluid would exert exactly the same pressure on any other body occupying the same space in the fluid. Hence the resultant vertical pressure on any body immersed is equal to the weight of the fluid displaced, and acts at its centre of gravity, which point is called the centre of displacement^ or centre of buoyancy. The principle thus established is commonly known as the principle of Archimedes ^"^ and may be thus enunciated : When a body is immersed in a fluid it is subject to a force equal to the weight of the fluid displcued^ which acts at the centre of buoyancy vertically upwards. Or, A body immersed in a fluid loses a portion of its weight equal to the weight of the fluid displaced. Before proceeding to consider some of the chief deductions from this proposition, we will show how it may be experimentally verified. § 45. Experiments. — i. Take a vessel witii a spout in one side, as shown in fig. 22. Pour in water till it Fig. 22. begins to run out from the spout. Take a piece of iron weighing i lb., and having suspended it by a fine thread from one of the scale-pans of the hydrostatic balance, weigh it in the water, allowing the water displaced to escape into another vessel, B. The piece of iron will ^ be found to weigh nearly 14 oz., and the weight of the water col- lected in the vessel b will be found to be a little * Born at Syracuse in Sicily, flourished about 250 B.C. Digitized by VjOOQ IC Principle of A rchimedes, 47 more than 2 oz., thus showing that the loss of weight of the body is equal to the weight of the water displaced. 2. Take a hollow brass cylinder, a, into which a solid cylinder, b, exactly fits. Hang the hollow cylinder to one of the Fig. 23. scale-pans of the hy- drostatic balance, and attach the solid cylinder to it by means of the hook, as shown in fig. 23. Now weigh carefully the two cylinders. Having observed their weight, take a vessel containing water, and let the solid cylinder hang in it completely immersed. Equi- librium is destroyed, and the free scale-pan de- scends. Now fill the hollow cylinder with water, and equilibrium is at once restored, clearly showing that the weight of the water in the hollow cylinder, i,e, the weight of a quantity of water of the same size as the body immersed, is equal to the loss of weight of the solid in water, Le, to the upward pressure which the liquid exerts on the body. § 46. Beal and apparent Weight of Bodies. — We have seen that a body surrounded by a fluid is pressed upwards by a force equal to the weight of the fluid displaced. This is the case with all bodies weighed in air, and consequently their real weight (i.e. their weight in vacuo) is greater than their ap- parent weight by the weight of the air displaced Digitized by VjOOQ IC 48 Hydrostatics. If two bodies, the volumes of which are not equal, balance one another in air, their real weights are not the same : that which has the greater volume is really the heavier. Thus, let the volumes of the two bodies be V and V\ their real weights W and W, Then, if they balance in air, W^ Vs=: W— Vs, where s is the specific gravity of air. Hence, W^W' + { F- V% or, ^is greater than W, if Fis greater than V. When the weights of two bodies, of small and nearly equal volumes, are compared, the diflference between their real and apparent weights is so slight as to be in most cases of no practical importance. But where the volume of one body is much greater than that of the other, the quantity ( F— V')s cannot so easily be neglected. Thus, in answer to the question. Which is the heavier, a pound of feathers or a pound of lead ? we should say that the apparent weights of both are the same ; but the real weight of the feathers is greater than that of the lead, and the two would not equilibrate each other in vacuo. § 47. Bzamples.— (i.) A solid cube of metal the edge of which is 3 inches, and whose specific gravity is 7, is wholly immersed in water and is supported by a string attached to it. Find its apparent weight in water. Let Wh^ the weight of the body in air, which differs very slightly from its real weight, A its apparent weight in 'wSiter, and y the resultant vertical pressure acting upwards. Then the body is in equilibrium under the action of these three forces, and, therefore, W^ Y+A, Now, the resultant vertical pressure K equals the weight of Digitized by VjOOQ IC Fluid-pressure on Bodies immersed, 49 water displaced «3» xze;, where w=» weight of a cubic inch of water, and W^ 3* x 7 x w. Hence -4« W— y«3*x 7xw-3»xw = 6x3»xz(/ 162 X 1000 ^ „ ^ - ° .1728 °'- = Slbs. i3|o^. (2.) A body weighs in vacuo 560 grams, and in water 60 grams : find its volume. Here W^ A = 560 - 60 = 500 grams a weight of water dis- placed. Hence volume of the water displaced, ue., the volume of the body = 500 cubic centimetres. {3.) Two hollow spheres, the volumes of which are 100 and 200 cubic centimetres respectively, balance one another in vacuo. What weight must be placed inside the larger that they may balance in -water ? Weighed in water the larger sphere will seem to be the lighter, since the force supporting it is greater. Now the force supporting the larger sphere in water is 200 grams, and the force supporting the smaller sphere is ic» grams. Hence, for equilibrium, the weight of the larger must be increased by 100 grams without increasing its volume. This may be done by placing 100 grams inside. {4.) Find the acceleration with which a heavy smooth body will sink in a perfect fluid less dense than itself. Let ff^-the weight of the body, 5 its absolute specific gravity. Let / = the specific gravity of the fluid. Then the volume of the body is — - (§ 17) and the weight of 5 the fluid displaced is W—, Hence, the resultant force measured in gravitatic*n units, causing the body to descend, is W ( i— -^ and the mass of the body moved is — . d by Google 50 Hydrostatics, Ifi therefore, / is the acceleration with which the body de- scends, it is shown in books on Mechanics that If the body be specifically lighter than the fluid, the body, if placed under the fluid, will tend to rise to the surface, and the acceleration /= ^-^ . g. s Exercises. V. 1. Find the weight in water of a piece of zinc (sp. gr. =7) that weighs 8 oz. in vacuo. ^ *" 2. Find the volume of a body that weighs 350 grams in vacuo and 225 in water. ' • u j ^ - T i J - liJ^t V^ 3. A block of stone 2 cubic feet is wholly immersed in . water. With what force is it buoyed up ? ". 3 • i '^ I 1 I Xf^tjb 4. A piece of metal weighs 36 lbs. in air and 32 lbs. in fresh water. What will it weigh in sea-water the sp. gr. of which is 1-025? y-<jvj-. V -- ^-z 5. An air-ball, the volume of which is 6 cubic decs., weighs in air 15 grams. Find its real weight, having given that the weight of I cubic centimetre of air is 0*0013 grams. 6. A round disc of lead (sp. gr. = 1 1 '35) area 5 sq. centi- metres and thickness 2 centimetres, is fixed to a piece of cork (sp. gr. =o*24) of same area and 8 centimetres thick. Find the weight of both in water. 7. The edge of a hollow cube of lead is 8 centimetres, its , tliickness is 2 centimetres. Find its weight in water. /'•^•*' ^^^ " '^''^ 8 A bottle weighing 200 grams is completely filled with water, when it is found to weigh 800 gr. If a piece of iron (sp, gr. =7*2) is placed in the bottle, the bottle with its contents weighs 955 grams. Find the weight of the iron. 9. A piece of metal of specific gravity 8, and weighing d by Google Fluid-pressure oft Bodies immersed. 5 1 20 lbs., is dropped into a cylinder Jilledmth water. Find the additional pressure on. the base. 10. A cylinder of wood (sp. gr. «=o*75), one decimetre high, with a sectional area of 16 sq. cm., is immersed with its axis vertical in alcohol (sp. gr. =o*8). Find the least weight that must be placed on the top of it to bring its upper surface on a level with the alcohol. 11. A piece of wood weighing 8*25 grams (sp. gr. bo*66) is placed 4 ft. 7 in. deep in water and is free to rise. N^lecting all frictional resistance, find its velocity when it reaches the sur- face of the water. 12. Find the time occupied by a stone (sp. gr. b3*2) in falling from rest through 55 feet of water. 13. Two spheres whose radii are respectively I^ and r cm., and both of which are heavier than their respective bulks of water, are of equal weight. What weight of metal (sp. gr. =j) must be attached to the doUom of the larger that they may balance each other in water? 14. Two masses of given specific gravities balance when suspended from the equal arms of a lever in a known fluid. What is the specific gravity of a fluid in which they balance when one of the masses is doubled ? 15. A cubic inch of one of t\ro liquids weighs a grains, and of the other 6 grains. A body immersed in the first weighs / grains and in the second g grains. What is its reai weight and what is its volume ? VIII. F/oafing Bodies — Metacentre, § 48. When a body is immersed in a fluid, we have to consider three separate cases : — I. The weight of the body may be greater than the weight of the fluid displaced, in which case motion will take place in the direction of the greater force, and the body if left to itself will sink. E 2 Digitized by Google 52 Hydrostatics, 2. The weight of the body may be equal to the weight of the fluid displaced, in which case it will rest anywhere in the fluid. 3. The weight of the body may be less than that of the fluid displaced, in which case the re- sultant vertical pressure will force the body upwards, and it will float These three cases can be easily illustrated by ex- periments. The first of them has been already con- sidered in the preceding paragraphs, and it has been shown that in order that the body may be prevented from sinking in the fluid, it must be upheld by a force A equal to its apparent weight, such that when ^is the weight of the body in vacuo, and Y the resultant vertical pressure, or weight of the fluid displaced. The second occurs less frequently, and does not now need to be separately treated. Later on, when speaking of the formation of drops, we shall have occasion to consider the form assumed by a mass of fluid which is wholly supported by the external re- sultant pressure. The third case is of great practical importance, and brings us to the consideration of floating bodies. § 49. Principle of Flotation.— If we take a body the weight of which is less than the weight of an equal volume of a liquid into which it is immersed, the resultant vertical pressure will bring it to the sur- face, and the body will be found to assume a posi- Digitized by VjOOQ IC Principle of Flotation. 53 tion of equilibrium in which it will be only partly immersed. If the experiment indicated in § 45 (i) be tried with a body lighter than Fig. 24. water, it will be found that the weight of the liquid that escapes in consequence of the partial im- mersion of the body is exactly l^^^^^^^Sf equal to the weight of the body ]|OI itself. ' ^ The relation between the part of the body which is immersed and the part that rises above the sur- face of the fluid is determined by the principle that the two forces acting on the body must equilibrate each other ; ue. the weight of the whole body acting downwards must equal the weight of the fluid dis- placed acting upwards. Thus if a b (fig. 24) be the intersection of the body with the surface of the liquid in which it floats, the position of a b is determined by the equation W =■ Y^ where ^ is the weight of the body, and Fthat of the liquid displaced Moreover, if G (fig. 24) be the centre of gravity of the body and s the centre of buoyancy, g and s must be in the same vertical line, for otherwise the body would be acted upon by a couple which would produce oscil- lation or rotation. Hence the conditipns of a body floating in perfect equilibrium are : — 1. The weight of the body must equal the weight of the fluid displaced. 2. The centres of gravity of the body and of the fluid displaced naust He in the same vertical line. § 50. Stable and unstable Equilibrium. — If a floating body be slightly displaced, so that the points Digitized by VjOOQ IC 54 Hydrostatics. s and G do not lie in the same vertical line, the mo- ment of the forces acting on the body may either tend to restore the body to its original position, as in fig. 25, in which case the equilibrium is said to be stable^ or to overturn the body altogether, as in fig. 26, in which case the equilibrium is unstable. Whether a floating body, partly immersed, can suflfer a small displacement without being overturned is a matter of the greatest practical importance, as the safety of all vessels at sea depends on it For, in consequence of the action of the tides and waves, a vessel seldom or never floats in perfect equiHbrium, but is contmually undergoing a rotatory displacement, which makes it oscillate about its original position of equilibrium Now we see that if s is above G the body will alv^ays return to its original position, or in this case the equilibrium of the vessel may be said to be stable. But where s is below G, the body is very likely to overturn if displaced, and hence the danger of overloading the deck, and the advantage of ballast in lowering the centre of gravity of the vessel. But a body may float in stable equilibrium even if s be below G, as we shall see presently ; and consequently -" ----- Digitized by VjOOQIC Stable and Unstable Equilibrium. 55 we cannot determine from the relative positions of s and G only whether the equilibrium of a floating body is unstable or not. § 51. Metacentre. — When a body floating in equi- librium receives a slight displacement, the centre of gravity of the fluid displaced assumes a new position on \^hich the stability of equilibrium mainly depends. Thus let ABC (fig. 27) be a body floating in equili- brium, and let s be the centre of buoyancy. Suppose, now, the body to undergo a slight displacement, in Fig. 37. consequence of which the centre of buoyancy changes to s'. If, then, the vertical through s' meets in the point M, the line through s and G which was vertical in the first position of the body, it is clear that when m is above g the equal forces acting on the body will form a couple tending to restore the body to its former position, and that when m is below G they will tend to overturn the body. The point m is called the metacentre of the body, and its position depends on the shape of the body and on the position of its centre of gravity. Of course, if g is below s the point m will be above g. Hence it follows generally that the equilibrium of a floating body is stable or unstable Digitized by VjOOQ IC 56 Hydrostatics, according as the metacentre is above or below the centre of gravity of the body. The metacentre may be defined as the point in which the vertical through the centre of buoyancy of a floating body which has undergone a slight dis- placement, intersects the line drawn vertically through the centre of buoyancy in the original position of equi- librium. If a body floats wholly immersed in a fluid, no change in the position of the body will alter the relative positions of the centres of gravity and buoy- ancy. Hence in this case the equilibrium cannot be stable unless the centre of gravity is below the centre of buoyancy. If the positions of these points be reversed, the slightest oscillation will cause the body to overturn. § 52. Bxamples.— (I.) A body the specific gravity of the material of which is s floats in a liquid specific gravity s'. Find what fraction of its volume will be immersed. Nearly all problems on floating bodies may be solved by the application of the principle of Archimedes, i.e, by equating the weight of the floating body with the weight of the fluid it dis- places. Let Fequal volume of the body; V of the portion immersed. Then K^ = weight of body; and F'/s: weight of fluid dis- placed ; V s (2.) A cylinder the height of which is 12 inches floats two- thirds immersed in water. Find what part of it will remain im- mersed if a liquid the specific gravity of which is o*2 be poured upon the surface of the water so as to completely cover the cylinder. Digitized by VjOOQ IC Fluid-pressure on Bodies immersed. 57 When the cylinder floats in the water only, the weight of the cylinder equals the weight of the water displaced. where s is the specific gravity and a the sectional area of the cylinder. ,*, specific gravity of cylinder = §. When the lighter liquid is poured on to the surface of the water, the weight of the cylinder equals the sum of the weights of the fluids displaced. If z equals in this case the part of the cylinder immersed in the lighter liquid we have I2xf = 2X0*2+(I2 — 2) or 80 = 22+ 120- IQ2 .', 2 a. 5 inches. Hence the cylinder rises one inch out of the water in conse- quence of the additional upward pressure due to the liquid poured into the vessel. (3.) Required the weight that must be attached to the end of a straight rod of known weight and specific gravity, that it may float vertically in water. A uniform straight rod cannot float in a vertical position, because the centre of gravity is always above the centre of buoyancy. If, however, a heavy Fig. 28. particle of no considerable magni- tude be attached to the lower end of the rod, the centre of gravity may be lowered as much as we please. What is required, therefore, is the weight of a particle that will lower the centre of gravity to the depth, at least, of the centre of buoyancy. Let G be the centre of gravity of the rod A b, a its sectional area, and s its absolute specific gravity, and lV\\s weight. Let x be the weight attached to the end A, and g be the centre of gravity of ^and x. Then w+a> d by Google 58 Hydrostatics, {IV+ x)kg=: JVx A G where ^= 2AGxaxs Also, by principle of Archimedes, IV-k- x^AK,a a Now, in order that the rod may float vertically, Ag must no be greater than A s. Put A^« A s. Then H^ _ W+x {IV+x)2as 2a .'. ^^{iv+xy This is the /easi weight that will suffice. Any greater weight than this may be added to A, provided it is not great enough to completely immerse the rod. Hence, the maximum weight possible is x, where — = IV +x, or x^fV (7-)- The weight required, therefore, must have a value some- where between ^(^_,)a„d^(i-,) In the above calculation the volume of the heavy particle, being P small, has been altogether neglected. (4. ) A uniform rod of given length and specific gravity moves freely about a point at a given height above the sur- face of a liquid in which it is partly immersed. Required its position of equilibrium, I>et A B be the rod ; PV, Y, G and s as before. The position of equilibrium is deter- "^^- d by Google Fluid-pressure on Bodies immersed. 59 mined by equating the moments of W and Y about A. Tak- ing the sectional area, which is uniform, as unity we have ff^=ABxjand y=BKx^, where s and d are the specific gravities of the material of the rod and of the liquid respectively. Also ^xGM=yxSN /. if=^JL Y GM . -iLuL^Af by ^imilaiity of triangles. BK. rf AG "^ ' ^ • AB S _ A B + AK _^AB ' ' AB-AK * d 2 * 2 •■•?-'-(^)' •••rfvo-j) •••— 'vo-a- If A c is given, A K can be expressed in terms of A c and of the angle B a c, which is thus determined. Exercises. VI. 1. A cylinder of wood (sp. gr. =0*6) and 12 inches high floats with its axis vertical in water. To what depth will it be immersed ? 2. A cube of oak (sp. gr. =097) each edge of which is 6 inches, floats partly in sea water (sp. gr. i '028) and partly in olive oil (sp. gr. =0-915). Find what part of it is immersed in each liquid. 3. A uniform block of metal 10 inches high (sp. gr. = 8) floats in mercury (sp. gr 13*6). Plnd how much it rises out of the mercury if water be poured on the surface of the mercury so as to completely cover the block of metal. 4. A cylinder floats vertically in a liquid. Compare the forces necessary to raise it and to depress it to an equal extent. 5. A heavy uniform rod weighing 3 kils. and sp. gr. = 6 moves freely under water in a vertical plane about a hinge at one end. If a string tied to the other end supports the rod in a horizontal position, find the tension in the string. Digitized by VjOOQ IC 6o Hydrostatics. 6. A piece of gold (sp. gr. 19*25) weighing 96*25 grams immersed in a vessel full of water causes 6 grams of water to be displaced. Is the gold solid ? If not, find the size of the hollow. 7. Two equal globes, the volume of each being 100 c.c, are suspended from the equal arms of a lever, the one hanging com- pletely immersed in water, the other in a liquid of sp. g^. =0*8. What additional weight is required to make them balance ? 8. A cube of metal is floating in mercury (sp. gr. = 13*6). When a weight of 170 lbs is placed on the top, it is observed to sink 3 inches. Find the size of the cube. 9. If an iceberg (sp. gr.= 0*918) float in sea-water, what is the ratio of the part submerged to that which is seen above water ? 10. Find what quantity of cork must be attached to a man whose weight is 168 lbs. and sp. gr. = i -12 so as to enable him just to float in water. 11. An iron ball of 12 lbs. weight floats in mercury covered in water. Find the weights of the parts in the two fluids ; having given specific gravity of mercury =13*6, specific gravity of iron = 7-5. 12. The specific gravities of the upper and lower of two fluids that do not mix are 0*9 and i*i ; the- upper fluid is 4 in. deep ; a cube with an edge of i foot and sp. gr. 0*75 floats in the liquids ; how much of it is immersed ? d by Google specific Gravity, 61 CHAPTER IV, SPECIFIC GRAVITY, AND MODES OF DETERMINING IT. IX. Application of the Principle of Archimedes to the determination of the Specific Gravity of Bodies, § 53. To find the specific gravity of a substance we require to know its weight in vacuo, and its volume, W since 7^ = ^f- The volume of the body, expressed m cubic centimetres, is numerically equal to the weight of an equal bulk of water at the standard tempera- ture expressed in grams, since the weight of a unit- volume of water is one gram ; and, therefore, the specific gravity of a substance is expressed numerically by the ratio of its weight to that of an equal bulk of water. Instead of the weight of the body in vacuo, we generally substitute the weight in air, the difference being unimportant for solid and liquid bodies of small magnitude. Where the volume of water equal to that of the body can be directly found, the specific gravity is very easily determined ; but where this is not the case different methods have to be employed. The principle of Archimedfts tells us that a body immersed in a fluid is pressed upwards bv a force Digitized-by VjOOQ IC 62 Hydrostatic!:. equal to the weight of the fluid displaced. Hence the specific gravity of a body can be found by comparing the real weight of a body with the difference between its real weight and its apparent weight in water, since this difference is equal to the weight of water dis- placed. If, therefore, W represent its real weight, and A its apparent weight in water, the specific W gravity of the body equals ^ . § 54. To find fhe specific gravity of a solid body insoluble in water. This can be determined roughly by an experiment similar to that described in § 45 (i), where the weight of the body may be directly compared with the weight of the water displaced. But a more accurate result may be obtained by using the hydrostatic balance for finding A^ the weight of the body in water. The specific gravity can then be determined by dividing W^ the weight of the body, by W^A^ the loss of weight in water ; or specific gravity = -^ — - % 55. To find the specific gravity of a solid body that fioats in water. In this case the body may be said to have a nega- tive weight in water : Le, it requires the application of a positive force to make it weigh zero when completely immersed. Let ^ equal the weight of the body, P the force required to completely immerse it, then the force necessary to resist the resultant vertical pressure = ^ + P = weight of water displaced by the whole body. Digitized by VjOOQ IC specific Gravity. 63 W Hence, specific gravity = W^-P' What we want, therefore, is to determine P, This we can do by attaching a heavy body called a sinker to the light body, and by weighing them both in water. If they weigh zero, the sinker's weight in water is the force P required ; but if they tend to sink and are found to weigh -5, then the sinker's weight in water is too great by this weight B. If, therefore, A equals the apparent weight of the sinker in water, A-^B^P, and the specific gravity of the body is r^jz :; =. fr + A — o For determining these weights the hydrostatic balance is again employed. § 56. To find the specific gravity of a liquid, by weighing a solid in it. Let the solid weigh W in air, and let a be its ap- parent weight in the given liquid. Then ^—dJ = weight of the volume of liquid displaced by the solid. Let solid weigh A in water. Then W— A^ weight of the volume of water displaced by the solid. But these two volumes are the same ; .•, specific gravity of liquid = " § 57. To find the specific gravity of a solid body soluble in water, but insoluble in a liquid of known specific gravity. Digitized by VjOOQ IC 64 Hydrostatics, If the body is soluble in water, its weight in water cannot be directly found, and consequently the method of § 54 is inapplicable to this case. If, however, the body is insoluble in some other liquid the specific gravity of which is known, we have the means of determining the specific gravity of the body. Suppose the weight of the body to be W^ and its weight in the known liquid a^ then W — a \% the weight of the liquid it displaces ; and if s be the spe- cific gravity of this liquid, — ^^ = the volume of • s liquid occupied by the body. It follows, therefore, W—a that the weight of an equal volume of water is , since specific gravity of water is unity, ^ W—a TVs Hence, specific gravit}' of body= JV^ — W-a § 58. Specific Oravity of Oases. — In determining the relative weight of gases, air at 0° C, and at the ordinary atmospheric pressure is taken as the standard substance. The process is attended with so many difficulties, owing to the peculiar properties of gases which have not yet been considered in this work, that many precautions are needed in order to obtain accurate results. In comparing the weights of equal volumes of any gas and air, it is necessary that the gases should be subjected to the same atmospheric pressure, and should be brought to the same temperatiure. If we suppose JV to be the weight of a glass globe full of air, and w its weight when empty, and if fV' Digitized by VjOOQ IC specific Gravity, 65 be the weight of the same globe filled with any other gas, then the specific gravity of this gas as compared with air is W— w If the globe be filled with water from which all air-bubbles have been carefully excluded, and if Wx be its weight so filled, the specific gravity of the gas as compared with water would be -— . ^ Wx—w In determining with accuracy the specific gravity of any substance, the temperature at which the experi- ment is conducted must be considered; for it is known that bodies generally expand when heated, and the weight of the same volume of water, or of any other substance, consequently varies at different tem- peratiures. This subject is more fully considered in treatises on heat Before working the following exercises the student is recommended to take pieces of brass, tin, zinc, marble, wood, and other easily obtained substances, and to determine their specific gravities by the hydro- static balance, comparing the results with those given in the tables. Exercises. VII. 1. A solid soluble in water but not in alcohol weighs 346 grams in air and 210 in alcohol. Find the specific gravity of the solid, that of alcohol being 0-85. 2. A solid weighs 100 grams in vacuo, 85 grams in water, and 88 grams in another fluid. What is the specific gravity of the fluid? F Digitized by VjOOQ IC 66 Hydrostatics. 3. A piece of copper weighs 31 grains in air and 27*5 in water. Find its specific gravity. 4. A piece of wood (sp. gr. ==074) of 32 cubic inclies floats in water. How much water will it displace ? 5. A body floats in water with one-eighth of its volume above the surface : determine its specific gravity. How much of it will be submerged in a fluid whose specific gravity is 09 ? 6. A body floats in a liquid (sp. gr. =si3*5), ^^^%. ^^ ^^s volume is above the surface : find specific gravity of the body. 7. If a ball of platinum weigh 21*4 oz. in air, 20*4 oz. in water, and 19-6 in hydric-sulphate, find the specific gravity of the platinum and of the acid. 8. A piece of marble, specific gravity = 2*84, weighs 92 grams in water and 98*5 grams in oil of turpentine. Find the specific gravity of the oil. 9. A piece of cork weighs 2 oz., and its specific gravity is 0-24. What is the least force that will just immerse it? 10. A piece of iron, sp. gr. 7*21, and weighing 360*5 grams is tied to a piece of wood weighing 300 grams, and the weight of both in water is 110*5 grams. Find the specific gravity of the wood. 11. A cubic block of wood each edge of which is 8 cm. is put into water. If the specific gravity of the wood is 0*85, with how much more wood must it be loaded, so that its upper sur- face may sink to the level of the water ? 12. A cylinder of wood 20 inches in height, and a cylinder of lead I inch in height, are united so as to form one cylinder 21 inches in height, which is found to float in water with 3 inches projecting above the surface. Find the specific gravity of the wood, that of the lead being 1 1 -4. X. Ot^er Methods of obtaining the Specific Gravity of Substances. — Hydrometers. § 59. The Specific Gravity Bottle.— This is a bottle made to hold a certain weight of water at the d by Google The Specific Gravity Bottle. 6^ standard temperature. It may be employed for deter- mining the specific gravity of a liquid or a powder. 1. Specific Gravity of a Liquid, — Suppose the bottle when empty to weigh w grams, and that it holds I GO grams of water. Let the bottle be filled with the liquid the specific gravity of which is required, and suppose it then to weigh (91+ a/) grams. Then the weight of the liquid in the bottle is 91 grams, and the weight of an equal bulk of water is 100 grams. Hence, specific gravity of liquid is -^ = 0*9 1. 2. Specific Gravity of a Powder, — Let the powder be first placed in the bottle, and let the bottle be then filled with water, and suppose the weight of the contents of the bottle, /.^., of the powder and water, to be K, Then if ^be the real weight of the powder, and y the weight (unknown) of the water it displaces, ^^=100+ w- y, and .-. Y^W-^-ioo-K W /. Specific gravity of powder = -j^ Take 4 grams of powdered glass, put it into the bottle, and fill with water ; then the bottle with its contents will be found to weigh 102*5 + 0/ grams ; /• F=4— 2-s=i-5; .•, specific gravity=Ttj=2-6. § 60. Hydrometers.— The hydrometer^ consists essentially of a straight stem, loaded at one end, so as * This instrament is said to have been invented by Hypatia, the daughter of Theon Alexandrinus, who flourished about the end of the fourth century ; though there is some foundation for the opinion that tht invention is due to Archimedes. F 2 Digitized by VjOOQ IC 68 Hydrostatics, to float in a vertical position. By observing the depths to which it sinks in two different Hquids, the relative weights of these two liquids can be deter- mined. Hydrometers are of very different forms, and may be used for finding the specific gravity of liquids or of solid bodies. We shall describe two varieties only. § 6i. Common Hydrometer. — ^This instrument, invented by Fahrenheit,* consists of a straight stem, Fig. 3a usually made of glass, which terminates in two ^ hollow spheres. The lower sphere is loaded ^ with mercury, so that the instrument may float in a vertical position. Let a = section of the stem. w = weight of the instrument. V = volume „ „ Suppose the instrument to float in water with * the point d of its stem on the surface of the water, and in some other liquid, the specific gravity of which is to be found, with the point c on the surface. Then, if j be the specific gravity of the liquid, it follows from the principle of Archimedes that 'w = si^—a X A c) = v—a x a d taking the weight of the unit-volume of water as unity. • __ Z^~gXAD § 62. Nicholson's Hydrometer.' — This form of instrument enables us to determine the specific gravity > 1724. * 1787. Digitized by VjOOQ IC Hydrometers, 69 of solid as well as of liquid bodies. It consists of a hollow body, b, connected by a fine stem with a small dish, A, at its upper end, and at its lower end with a cup, c, loaded so as to ensure equilibrium. When floating in distilled water, with a certain weight, k^ in the upper dish, the instrument sinks to such a depth that a fixed mark, 0^ is on the surface of the water. 1. If we want to find the specific gravity of a given liquid, we place the instrument in it, and place weights in the upper dish ^^^' 3'' till the fixed mark, ^, is on a level with the surface of the liquid. If, then, W be the weight of the instrument, k the weight added to sink it to in water, and w be the weight added to sink it to in the given liquid, W-^-k is the weight of the water dis- placed ; and W-^-w is the weight of the liquid displaced ; and, as the same bulk of fluid is displaced in each case, the specific gravity of the liquid is W+k' 2. To find the specific gravity of a solid. Let k be the weight as before that must be placed on the instrument to bring the point ^ on a level with the surface of the water. Place a small piece of the solid on a, and diminish the weight k^ so as to keep the instrument at the same level. Let m be the weight now employed. Then k—m=si weight of solid body. Digitized by VjOOQ IC yo Hydrostatics. Now place the substance the specific gravity of which is to be found in the cup c, in which case a weight greater than m by the weight of the water displaced by the body must be placed on a to main- tain the instrument in the same position. Let n equal the weight now in a. Then «—f« = the resultant vertical pressure on the solid in c, = the weight of the water displaced by it Hdnce, specific gravity is -— ^. Exercises. VIII. 1. A piece of glass the weight of which is 50 grains is placed on the upper dish of a Nicholson's hydrometer, and it is found that an additional weight of 235 grains is required to sink the instrument to the fixed level. If, however, the glass be placed in the lower cup a weight of 250 grains is required. Determine the specific gravity of the glass. 2. A globe of glass holds 5 litres of water at standard tem- perature. When full of water it weighs 5500 grams. "When full of air it weighs 506 "465 grams, and when full of hydrogen it weighs 500*447 grams. Find the specific gravity of air as compared with water, and of hydrogen as compared with air. 3. A bottle filled with water is found to weigh 500 grams. If 180 grams of powder are introduced into the bottle, the bottle with its contents weighs 575 grams. Required the specific gravity of the powder. 4. Into a specific gravity bottle capable of holding looo grains of water, 300 grains of a certain powder are introduced. The bottle is then filled up with a liquid, specific gravity o*8, and the contents of the bottle are found to weigh 980 grains. Find the specific gravity of the powddr. d by Google The Motion of Liquids. 7 1 CHAPTER V. THE MOTION OF LIQUIDS. XI. Liquids moving by their own weight, § 63. It is proposed to consider in this chapter a few of the most elementary propositions connected with the movements of fluid bodies. Very little can be attempted without the application of some of the higher processes of mathematics, and we shall endea- vour, therefore, only to indicate the principles of the methods on which the solution of this class of pro- blems depends. As all liquids are more or less viscous, absolute agreement cannot be expected to exist between the conclusions arrived at on the assumption of a frictionless fluid and the results of direct ex- periments. § 64. DefinitioD. — The vertical distance between the surface level of a liquid in a vessel and the centre of the orifice through which it escapes is called the head ox pressure height under which the flow takes place. § 65. Torricelli's Theorem. — The velocity with which a liquid escapes from a small hole in the bottom or side of a vessel was the subject of numerous experi- ments made by TorricellL^ The result at which he * Born in Italy in 1608 ; died in 1647. Digitized by VjOOQ IC 72 Hydrostatics. arrived was that the rate of efflux was the same as would be acquired by a body falling freely from the surface level of the liquid to the centre of the orifice from which it escapes. Let A B c D be a vessel having a small open- ing, o, in one of its sides. Let the area of this open- Fig. 32. ing, which is supposed to be very small, be ^, and suppose a small mass of liquid, ;;/, the thickness of which is Cy to occupy this opening. Then if / be the pressure in excess of the atmospheric pressure urging this mass forwards, measured in units of force per unit-area, p=^g. da h where d is the density of the liquid, and h the pressure height, o H ; and the work done by this force p in urging the mass m from a position of rest through the distance c is p X c =i g, dah.c) and since the work done is equal to the energy which the mass m acquires, p , c '=' , where v is velocity of efflux ; sr.d ahc=i dac — ^ 2 or v^ -=■ 2 g h />., the velocity is that which would be acquired by the mass m in falling freely from the surface level of the liquid to the centre of the orifice. d by Google TorricellVs Theorem, 73 In this result each particle of the mass is supposed to move at right angles to the area of the orifice, and to escape into the air at a pressure equal to that at the surface level § 66. Experimental Test.— The accuracy of the results given by Torricelli's theorem may be tested for different liquids, by fig. 33. taking a vessel having apertures in one of its sides through any of which the water or other liquid it contains may escape. If the vessel be kept constantly full, and one of these aper- tures be opened, the « 6 c liquid will flow through it, and will descend in a curve similar to the path of a body projected horizontally from a certain height above the ground. If the distances payjf>b,pche observed, the accu- racy of Torricelli's theorem can be tested. For, let the height o c = ^, then according to the theorem, if v be the velocity of the liquid at c, V = -^ 2gh, and the velocity will be uniform and horizontal. If then / equal the time occupied by a particle in moving from c to c, we know by the second law of motion that / equals time occupied by a body falling freely from c to / = time of describing/^ with the uniform velocity v. Digitized by VjOOQ IC 74 Hydrostatics. Hence, pc^='tv\ also cp = — ^ orvj=pcx^/ S ^ V 2C/ Now, if the distance/ c be known, and if /^ be ob- served, the value of z/ so found can be compared with that given by the formula z/ = n/ (2^. o c). On comparing the results, obtained by experiment, with those obtained from Torricelli's theorem, a certain difference will be always found to exist. Some of the reasons of this difference, apart from those depen- dent on the viscosity of the liquid, will be considered later on. § 67. Belation of Telocity of Flow to Sectioiial Area of Vessel* — Let a b c d be a vessel, which is Fig. 34. kept constantly full of liquid. We may consider the mass of the fluid, which is supposed to be incom- pressible, to be divided into small laminse moving Digitized by VjOOQ IC Vena Contracta. 75 parallel to one another and of uniform sectional area for very small differences of depth. Thus, libcde, p^rsyhe two small laminae, we may suppose bc^=^de and/^ ^^ r s. Let each particle in the section be have the velocity, Vy and each particle in the section pq the velocity v'. If, then, the section be descend to d e in one second, the section / q will descend to some depth, rs, in the same time ; and the volume bcde will be equal to the volume /^rj, since the quantity of liquid between de and /^ remains con- stant If, then, A equal the section be ox de, and a' the section/^ or r J, A z/ = a' z/' = volume discharged in one second, /. V \ v' \\ a! : A, or the velocities are inversely proportional to the sectional areas. § 68. Vena Contracta.— When a liquid issues into the air from a small opening in a thin plate, the stream is found, first of all, to converge, so that it contracts rapidly for some little distance from the orifice. This fact may be very easily observed, and will be seen in all cases such as those shown in figs. 33, 34. The area of the jet at its narrowest part is known as the vena contracta^ and was first made the subject of investigation by Sir Isaac Newton. It is found to be generally about three-fifths of the aperture of the orifice, but varies with the shape of the aperture and the pressure height. The value of this fraction, which is called the eoefficient of eonir action ^ can be determined by experiments only. In all calculations with respect to the rate of the emptying of vessels, the method Digitized by VjOOQ IC jS Hydrostatics. usually adopted has been to apply the results of Tor- ricelli's theorem, and to substitute the area of the vena contracta for that of the orifice. This method, we shall see, gives us a merely empirical result, and is faulty in so far as it overlooks some important elements in the problem. § 69. MeasTire of Pressure on the Walls of a Pipe. — Suppose a liquid is flowing through a pipe A B, the pressure which it exerts on the walls of the pipe at any point can be experimentally determined by in- serting gauge glasses, />., thin glass tubes of about Fig. 35. f of an inch diameter, into the pipe at those points at which the pressure is to be found. Since fluids transmit their pressure equally in all directions, the forward pressure exerted by the moving liquid will be equally exerted on the walls of the pipe, and will force the liquid up the gauge glasses to a height corre- sponding to the pressure produced. If the pipe is of uniform sectional area the pressure produced at all points will be the same, friction being neglected, and the liquid in the pipe will rise in the gauge glasses to the same height at all points in the pipe. § 70. Eolation of Pressure to Sectional Area of Pipe. — If the sectional area of the pipe varies, the Digitized by VjOOQ IC The Motion of Liquids through Pipes. Jj pressure is no longer uniform, but is found to be greatest where the area is greatest, and vice versd. Suppose the liquid to be moving with a uniform velocity in that part of the pipe a b (fig. 35) which has a uniform sectional area. If, then, we consider an element a of the liquid, we see that the pressure on either side of a must be the same, since any increase of pressure from behind would cause it to move with an acceleration, which is not supposed to be the case. Suppose now the element to have reached b, and to be entering the wider part of the pipe, its velocity in the direction of the pipe's length will now be less than it was before, and will continue to decrease as the pipe widens. Hence the pressure of the liquid behind the element a urging it on must ever be less than that in front of it resisting its advance, and con- sequently the pressure must increase with the area of the pipe and with the decrease in the velocity of motion. After passing c, the widest part of the pipe, the velocity begins again to increase, and as the resist- ance to the forward motion of the particles must consequently have become less, the pressure is also diminished. These results can be practically illustrated by observing the height of the liquid in the several gauge glasses placed at different points in the pipe. Due allowance must be made for the frictional resistance which causes the liquid to rise to a less height in the glasses than it otherwise would, and gives a difference between the theoretical and actual results which in- Digitized by VjOOQ IC 7S Hydrostatics. creases with the length of pipe traversed. This re- sistance being neglected, we see that the pressure which a liquid exerts on the sides of a pipe through which it is passing varies directly with the sectional area, the velocity at the two ends of the pipe being the same. If/i and/z be the pressure exerted at points where the area of the pipe is a^ and azy we have A :/2::«i :«2- § 71. Belation of Velocity of Flow to the Pres- sure produced. — We have seen that in the case of a steady flow through a pipe of varying area, the velocity at different points varies inversely with the area, fric- tion being neglected. We have now to consider how the pressure changes with the velocity. This may be experimentally exhibited by fixing the larger end of a tapering pipe into a vessel of water kept constantly full, the upper surface having only the ordinary at- mospheric pressure upon it. If the pipe be furnished FiQ. 36. with gauge glasses, the pressure at different points can be observed. The hydrostatic pressure at a is evidently zero, or the same as that of the air into Digitized by VjOOQ IC The Motion of Liquids. 79 which it issues, but the liquid will be found to stand at a continually higher level in the gauge glasses as the pipe widens towards the vessel. The difference between the surface level of the water in any one of the gauge glasses and that of the vessel is called the fall of free level dX that point, and this fall is equal to the difference between the height of the column of liquid representing the hydrostatic pressure at a point in the stream, and the pressure-height at the same point, supposing the liquid to be in equilibrium. Consider now a small mass m of the liquid in the section of the stream at any point b, and let the section of this mass be equal to a unit of area, and its length equal to ^, so that it contains c units of volume. Then if/ be the actual pressure in units of force per unit of area at b, and if P be the pressure at the same point due to the depth of the mass m below the free surface l l' d of the liquid, then the whole amount of work which the mass m has received from the pressure behind it in moving through its own length from a position of rest is -Px r, and since / is the pressure it is exerting, the work which it gives to the liquid in front of it is/ x c. Hence the excess of work which it receives is {P—p) c. Now if JjTis the vertical depth of m below l l', and h the height of the liquid in the gauge glass above niy P=:gHdzxidp — gdh, where d is the density of the liquid, .-. {P^p)c^gd{H^h)c. But the amount of work gained by the mask m \^ equal to its energy. Digitized by VjOOQ IC 8o Hydrostatics. Hence, g d^H—h) c = ^^, where v is the velocity of mass m. And since m = dCy we have Z;2 = 2g {H—h) ^ 2gDG or, the velocity at any point in the stream is that due to the difference between the statical and actual pres> sure-height of the moving particles. Hence, m steady flow the velocity generated from rest is that due to the fall of free level. If we suppose the sectional area at b to be twice that at A, and at c twice that at b, and so on, then if V be the velocity at a, the velocity at b = ^, since the velocity varies inversely with the area in cases of steady flow \ and since z/^ = 2 ^ . a d, we have — =• 2^.dg, orBG = ^AD; 4 4 ' ' and — = 2 ^. D H, or c h=— ^ a d, and so 16 16 on, which gives the relation between the actual hydro- static pressure at a point, and the velocity with which the stream is passing that point § 72. Application to the Flow of a Liquid through an Orifice in the Base or Side of a Vessel.— When a liquid flows through an orifice, as previously con- sidered in the case of Torricelli's theorem, the flow does not actually take place in a direction at right Digitized by VjOOQ IC Flcnv of Water through Small Orifice, 8i angles to the orifice, at all points of the orifice, but va directions such as are shown in fig. 37. The nature of the flow is therefore ^'^' 37' somewhat similar to that indi- cated in the preceding section, in which the varying area of the pipe is represented by the space included between the several stream lines. At all points along the margin of the orifice the direction of the motion is tangential to the plane of the orifice, and the water comes at once into contact with the atmosphere. But at all intermediate points the issuing stream exerts a pressure greater than that due to the atmosphere, and consequently the velocity at any of these points is not that due to the height of the free surface of the liquid in the vessel, but is that due to the difference between this height and that of the column of liquid corresponding to the actual pressure at the particular point in the moving stream. This difference of surface level has already (§ 71) been referred to as the fall of free level ; hence the velocity of any particle in the issuing stream is that due to the fall of the free level \ or, if h equals the depth of a particle in the stream below the free surface of the liquid, and z equals the height of the column of liquid represent- ing the pressure at that particular point, then if v be the velocity with which this particle is moving, (I = s/zglh—z). In calculating the actual discharge in a given time Digitized by VjOOQ IC 82 Hydrostatics. through a small orifice in the base or side of a vessel, the following principles, which are essential conditions of the flow, have to be considered : 1. The absolute velocity at any point in the plane of the orifice is not that due to the vertical distance between the point and the free surface level of the liquid, since everywhere throughout the area, except along the boundary of the orifice, the liquid is under a pressure greater than that of the atmos- phere, and consequently, as we have shown above, the velocity is correspondingly diminished, or V s= \/ 2 g(/i — z), 2. If the orifice be divided into small horizontal bands, we cannot suppose the velocity to be constant throughout each band, since the direction of the motion of the fluid is different at different parts. 3. If we consider any vertical section of the liquid as shown in the figure, since the direction of the mo- tion towards the orifice is nowhere perpendicular td the orifice, except at its centre, the actual velocity of efflux is only the component of the whole velocity as determined by the first of these principles, acting in a direction at right angles to the plane of the orifice. These considerations show how very complicated is the problem of determining the actual discharge o a liquid, per unit of time, through a given orifice, and serve to indicate some of the causes of the discrepan- cies between the results of actual experiments and those given by Torricelli's theorem. Ordinarily, when the orifice is small, the discharge is roughly calculated by considering the velocity of efflux to be that due to the distance between the centre of the orifice and the Digitized by VjOOQ IC Flow of Water through Small Orifice. 83 level of the free surface, and by then substituting the area of the vena contracta for that of the orifice. Thus if a be area of a very small orifice, and y the co- efficient of contraction, and if h be the vertical dis- tance of the centre of the orifice from the free sur- face level, the discharge per unit of time is taken as y a ^/2 g h. But this result is not only a mere empirical result, the degree of accuracy of which de- pends on the value of y as experimentally determined, but it leaves out of consideration some important cir- cumstances under which the flow actually takes place. ^ § 73. Effect of Friction on the Pressure of a Liquid Flowing Vniformly througli a Pipe.— Hitherto we have taken no account of the frictional resistance which retards the passage of a liquid through a pipe, and which, apart from all other circumstances, causes the velocity of efflux to be less than that due to the entire pressure-height above the orifice from which the liquid escapes into the air. The effect of this frictional resistance on the pres- sure exerted by a liquid in flowing through a pipe of uniform sectional area may be conveniently illustrated by the following arrangement. H A (fig. 38) is the section of a cylindrical zinc vessel about 3 ft. high, into the side of which, near the bottom, is fixed a brass pipe of uniform bore. At equal dis- tances along this pipe are openings into which glass gauge-glasses are fitted. If the opening e be closed, and the apparatus filled with water, the liquid will stand at the same > *0n the Flow of Water through Orifices.* By Prof. James Thomson. Report of British Association, Glasgow, 1 8 761 Digitized by VjOOQ IC 84 Hydrostatics. level in the vessel and in the pressure-tubes ; but as soon as the end e of the tube is opened, the water falls to different levels in the several tubes, being lowest in the tube nearest the end, and rising by equal differences in the several tubes. If the water in the vessel be kept at the same level, it will be observed Fig. 38. that the surface levels of the liquid in the several tubes lie on a straight line drawn through e to some point e'. This line is found to be more nearly horizontal, the greater the sectional area of the pipe through which the water flows. Now it is evident that if the liquid diiring its flow had encountered no frictional resistance, its velocity on leaving the vessel a h would have been that due to the pressure-height above it, and its pressure on the tube would not have exceeded that of the atmosphere into which it issued. The additional pressure on the tube is due, therefore, to the frictional resistance encountered by the liquid ; and as the resistance to be overcome at any point of the tube increases with the distance of that point from the orifice, the pressure of the liquid diminishes as it approaches the open end. Let a equal the sectional area of the pipe, and let c be Digitized by VjOOQ IC Effect of Friction. 8 5 the length of an element of liquid occupying it Let/ be the measure of the frictional resistance per unit of length of pipe, and let p and/' be the actual pressures per unit area at either end of the element c. Then if we suppose m to be the mass of this element, and %f its velocity, which is constant, — represents the 2 energy of this mass at all points in the pipe. Now if the mass m moves through a distance r, the work which it has received \^ pa x ^, and the work which it has given to the liquid in front of it \^p' a x c. Hence the resultant work gained in moving through the space c\^(p a — p' a) c. But since the energy of the mass remains constant, this amount of work is em- ployed in overcoming the frictional resistance encoun- tered by the liquid in its passage through the pipe. The work done against friction is/ x c\ and .% (p—p')ac^fy.x\ and since / — Z' ^=^gd(h—h') where h^ h' are the heights of water required to pro- duce the pressures / and ^ respectively, and d is density of the liquid, we have gda which shows that the difference of pressures (as mea- sured by height of the liquid in pressure-tubes) for equal distances along a pipe of uniform sectional area is constant. It is to be here observed that whilst the kinetic energy of the flowing water is constant, there is a continual decrease of potential energy much in the same way as when a body slides with a uniform velocity down a rough inclined plane* zed by Google 86 Hydrostatics, XII. Capillarity. § 74. In the preceding sections we have treated of the equilibrium and motion of liquids in mass, but we shall now consider a number of phenomena which are mainly due to the action of forces on the molecules of a liquid, and which are not immediately explicable by the principles already stated These will be dis- cussed under the two heads of Capillarity and Diffusion, and they comprise the formation of drops, the relation of liquids and solids in contact, and the intermixture of liquids, either in contact with one another, or separated by porous membranes. In discussing these subjects we shall in each case commence with a few experiments, and then indicate the nature of the prin- ciples which seem to explain them. § 75. Drop Formation. — ^Experiments.— Take a vessel containing olive oil and carefully drop into it some small portions of water. These will be seen to assume a spherical shape as they descend slowly through the oil Raindrops are spherical; but the rapidity with which they fall through the air is so great that we are unable to see them long enough to distinguish their form. To show that a large amount of liquid will assume a spherical shape if left free to the action of its inter- nal forces, mix together alcohol and water, in such proportions that the mixture may have tho density of olive oil This will require about three-parts of alco- hol to one of water. Then pour gently, so that it may not be broken into parts, some olive oil into the Digitized by VjOOQ IC Formation of Drops, 87 mixture, and it will be found to float in the form of a sphere in any part of the surrounding liquid. With^ out any great difficulty globules of oil may in this way be obtained, having a diameter of four or five inches. These can be flattened between two glass plates, or receive an indentation by being touched with a glass rod, without losing their cohesion ; and as soon as the external body is removed, they recover like an elastic ball, their spherical form. If pure quicksilver be scattered on a level surface of glass, the smallest of the drops will differ very little in form from perfect spheres. The larger drops, owing to their weight, will be considerably flattened ; and if a greater quantity cohere together, the tendency to a spherical shape is visible, only in the somewhat curved form of the upper surface. In blowing an ordinary soap-bubble we obtain a very good idea of the nature of the forces that act on a fluid surface. For this purpose * take some common soapsuds, or a Plateau's mixture of soap and glycerine, and blow a small bubble at the end of a tube with a bell mouth. A tobacco-pipe will answer if the bore of the tube is large enough. After blowing the bubble at one end of the tube, place the other end near the flame of a candle. The bubble will contract and drive a current of air through the tube, as may be seen by its effect on the flame. *This shows that the bubble presses on the air within it, and is like an elastic bag. To enlarge the bubble by blowing into the tube, work must be done to force the air in, because the pressure inside the bubble is greater than that of the air outside. This Digitized by VjOOQ IC 88 Hydrostatics, work is stored up in the film of the soapsud, for it is able to blow the air out again with equal force.' § 76. Surface-Tenrion.— -These experiments show that when two fluids are in contact with each other and do not mix, the surface separating them is in a state of tension similar to that of a membrane stretched equally in all directions, and that the surface-particles have a tendency to approach one another like those of the outer side of a bent watch- spring. This contractile force, or surface-tension, resides in the thin film separating the two fluids, and does not extend to any appreciable depth below the surface ; and to its action is due the spherical form of the drop and of the soap-bubble. In the case of the soap-bubble the superficial tension may be mea- sured by considering the work done in order to pro- duce a film of a certain area \ and the measure of the tension per unit length is found to be the same as the numerical value of the work done divided by the area.* This tension seems to be due to the fact that whilst the particles in the interior of a fluid are sub- jected to equal forces on all sides, those on the sur- face separating the two fluids are under the influence of different molecular attractions, the result of which is to give to the bounding surface a tendency to con- tract. The surface-tension depends on the nature and temperature of both media, decreasing as the temperature rises, and vanishing altogether at that temperature at which there is no longer a well- marked distinction between the liquid and gaseous states.^ Between any two liquids that do not mix, > C. Maxwell, Theory of Heat, pp. 281, 283. - § 4. Digitized by VjOOQ IC Surface-tension, 89 between a liquid and its own or another vapour, and between a solid and a fluid of any kind, there is a definite surface-tension, the value of whigh for a unit of length is called the co-efficient of superfi- cial tension, or co-efficient of capillarity, as having been first considered in connection with the ascent of liquids in capillary tubes. This tension may be reckoned in c. g. s. units of force per linear centimetre, and it is found that at a temperature of 2o°C. the tension of a surface of water in contact with air is 81, that of mercury and air 540, that of alcohol 25*5, and of olive oil 36'9 units of force. In fact, of ordi- nary liquids water is found to have the greatest sur- face-tension. § 77. Capillary Elevation and Depression. — Experiments. — (i.) If a sheet of clean glass be im- mersed in water and then removed, the water will be found to have wet the surface of the glass. If, whilst partially immersed, the surface of the water in contact with the glass be observed, it will be found that the surface of the water in the immediate neighbourhood of the sheet of glass is raised. If we perform the same experiment with mercury instead of water, we „ shall find that the mercury Fig. 39. Fig. 4a , , , -^ does not wet the glass, and that the surface of the mercury in the neigh- bourhood of the glass will be depressed. If A B be the surface of the liquid, c d the vertical section of the plate of glass, the water (fig. 39) will be found to rise to some Digitized by VjOOQ IC 90 Hydrostatics, height a b^ and the mercury (fig. 40) to be depressed to the level a' b\ Moreover, the surfece of the water in contact with the glass will be concave, and that of the mercury convex \ and the elevation of the water will be found to be somewhat greater than the depression of the mercury. (2.) Take now two plates of glass and let them ap- proach each other in water. When they are suffi- ciently close, the water will begin to rise between them, Fig. 41. and the height to which it rises will increase as the distance between them lessens. Moreover, the sur- face of water between the plates will be concave. If the same experiment be made with mercury instead of water, the mercury will be depressed between the plates, but to a less extent, and the surface will be convex. If this experiment be varied by taking plates of different thickness, the same results will be obtained, thereby showing that whilst the amount of ascent or depression depends on the distance between the plates, it is independent of their thickness. Digitized by VjOOQIC Capillary Phenomena. 91 Fig. 42. (3.) If instead of plain surfeces parallel to one another we take two plates of glass inclined at an angle, and immerse them partly in water, the water will rise be- tween the plates to different heights, and the intersection of the surface of the water with the plate will form a curve, which is known as the rectangular hyperbola. (4.) If glass tubes of small bore, called capillary tubes (from capiila, a hair), be partially immersed in water, the liquid will rise in the tubes, and the height to which it reaches will be found to vary inversely with the diameter of the tube. This result holds good whether the experiment be made in air or in a vacuum, but the elevation is foimd to diminish as the tempera- ture of the water increases. If mercury be used instead of water, the liquid will be depressed, as pre- viously stated in the case of parallel plates. The curved surface of the liquid is called a meniscus^ and the meniscus is concave for water and convex for mercury. § 78. Principles of Capillarity.— When a liquid comes in contact with a solid, the particles of the liquid on the common surface of the two substances are to some extent in a similar condition to those of the free surface of a liquid ; />., they are not equally attracted on all sides, as is the case with particles in the interior. The interaction of the molecular forces of the surface of the liquid and solid d by Google 92 Hydrostatics, gives rise to a tension in the common surface that separates them. When a solid is in contact with two fluids there is a different tension in each of the surfaces separating a pair of media. In this case, in order that the three tensions acting at any point in the Hne of intersection of the surfaces may be in equilibrium, their directions must be such that they can be represented by the sides of a triangle, and any one must be greater than the difference between the other two. Now we have seen that the surface of a liquid in contact with a solid is curved, and for any given liquid in contact with a given solid there is a definite angle of contact, called the angle of capillarity^ which is cunite in the case of mercury, and obtuse in the case of water in contact with glass. This inclination of the surface of a liquid to that of a solid at the margin of contact is due to the excess of the tension of the surface separating the two fluids (for example, air and water, or air and mercury) above the difference of the tensions of the surfaces separating the solid from each of the fluids. § 79. Height of ElevatioiL or DepressioiL of Liquid in Capillary Tubes. — Let T denote the mag- nitude of the superficial tension per unit length of the free surface of the liquid inclined at a given angle to the solid in contact with it. Then the vertical component of T acting upwards or downwards is the force tending to raise or depress the liquid. Let / be the vertical component of T', and let r be the radius of the tube, then 2 xr/ denotes the magni- tude of the vertical force acting all round the tube. If, now, s be the specific gravity of the liquid, h the mean height to which it is raised or depressed, then Digitized by VjOOQ IC Law of Diameters. 93 we have nt^hs equal to the weight of liquid supported, and 27rr/=7rr^^j, ox h = — rs which gives the law of diameters ^ viz., that the height of elevation or depression varies inversely with the radius of the tube. Since the surface-tension acts uniformly through- out the film, it produces a resultant pressure normal to the surface, which is always directed towards the centre of curvature of the surface. It follows, therefore, that in capillary elevation the liquid surface must be concave, and in capillary de- pression convex. In the case of a narrow tube partly immersed in a liquid which is in contact with air on both sides of the tube, we see that when the surface of the liquid within the tube is concave, the pressure immediately below the surface is less than the atmospheric pressure by a force due to the concavity of the surface, and that the pres- sure goes on increasing till at a depth corresponding to the mean level of the external hquid it is equal to the atmospheric pressure. If the surface is convex, the pressure immediately below the surface of the liquid in the tube is greater than the atmospheric pressure by a force corresponding to the difference of the mean level of the liquid inside and outside of the tube. Thus, in the case of a concave meniscus the resultant normal force due to the surface-tension acts iii a direction opposite to gravity, and so lessens its effect, and in the convex meniscus it increases it. Digitized by VjOOQ IC 94 Hydrostatics. Xlli. Diffusion of Liquids, § 80. DiffasioiL — If two liquids of different den- sities, and susceptible of permanent admixture, be placed in contact with each other, they gradually become intermixed. The process by which these liquids combine is known as diffusion, § 8t. Oraliam's Experiments. — The phenomena of diffusion were first carefully investigated by Pro- fessor Graham, and the results of his early experiments were published in 1850. In conducting these experi- ments Graham employed a number of small phials, each holding about 114 cc of water. The necks were carefully ground and of a uniform aperture of about 3*15 c. in diameter. Into these '^* ^^' ^ phials he poured solutions of various salts, so that the liquid rose as far as the shoulder of each phial. The phials were then very carefully filled with cold water. Thus charged, each phial was closed by a glass plate and then placed in a glass jar, containing sufficient water to cover the mouth of the phial and to rise at least 2*5 c above it. The plate was then carefully removed and the whole apparatus main- tained for a certain period of time at a fixed tempe- rature. The phial was then withdrawn and the water in the jar having been evaporated, the amount of salt that had passed from the phial into the jar by this process of diffusion was determined by weight. A very simple experiment showing the diffusion d by Google Diffusion of Liquids, 95 of liquids is the following : — ^Take a tall cylindrical jar and fill it to about two-thirds of its fig. 44. height with a solution of litmus. Then^ by means of a funnel pour in very care- fully some hydric-sulphate, so as to occupy the lower part of the jar. If the jar be set aside, the acid will be found, after the lapse of two or three days, to have diffused into the litmus solution, as shown by the consequent red colour it will have acquired. § 82. Besnlts of Graham's Experi- ments. — By varying the experiments with '\ respect to the nature of the salt, the den- sity and temperature of the solution, and the time occupied in diffusion, Graham arrived at the following results : 1. The increase in density in the diffusion product corresponds very nearly with the proportion of salt in the solution. Taking four different solutions containing one, two, three, and four parts of common salt to 100 parts of water, it was found that the quantities diffused at the end of equal times were very nearly in the pro- portion of I : 2 : 3 : 4, the variation from this result not exceeding one per cent. 2. The quantity of salt diffused in equal times in- creases with the temperature. From an experiment with a 4 per cent, solution, it was found that with a rise of temperature of i5°C., the quantity of salt diffused increased somewhat more than one-third Digitized by VjOOQ IC 96 Hydrostatics. 3. The rate of diffusion for weak solutions of the same salt is nearly uniform, but varies considerably with the nature of the substance diffused. Of the several substances employed in these experiments hydric-chloride was found to be the most diffusible, and albumen one of the least diffusible. Intermediate between these was common salt Com- paring common salt with sugar-candy and albumen, it was found that with solution of 20 parts of the solid substance in 100 parts of water, exposed for eight days at a temperature of 16° C, the specific gravity of the diffused solutions were respectively i'i26 ; 1*070 ; and 1-053. The following table shows the approximate times of diffusion of equal weights of different substances, taking that of hydric-chloride as unity : Hydric-Chloride. . . . .1 Sodic- Chloride . . . •2*33 Cane Sugar. 7 Magnesic Sulphate. ... 7 Albumen 49 Caramel 98 4. If a solution be taken of two different salts which do not chemically combine, it is found that each follows its own rate of diffusion. Thus the inequality of diffusion of two different salts supplies a method for their separation, to a certain extent, from each other. If a mixed solution of two corresponding salts of potash and soda be placed in the phial, the potash salt being more diffusive than the soda salt Digitized by VjOOQ IC Graham's Experiments, 9/ will escape into the water outside, whilst the soda salt will be relatively concentrated in the phial. 5. It is also found, in the case of very weak solu- tions, that a salt will diffuse into water which already contains some other salt in solution, showing that the diffusion of one salt is not sensibly resisted by the pre- sence of another. § 2^^, CrystalloidB and Colloids. — By consider ing the different diffusibiHty of different substances, Graham found that all bodies might be referred to two classes, which he called crystalloids and col- loids. The properties of these two classes of sub- stances are very different. Bodies susceptible ol crystallisation belong to the former class. They are highly sapid ; they generally form a solution which is very slightly viscous, and they diffuse rapidly through water, or through a porous diaphragm. Colloids, on the other hand (so-called from KoXXiy, glue), to which class belong starch, gum, hydrated alumina, albumen, gela- tine, and many organic compounds, are insipid and have very feeble chemical relations. They diffuse very slowly in water ; but they form a medium, like water, which arrests the passage of other colloids, but through which crystalloid substances are capable of diffus- ing. A peculiar property of these substances is their mutability. They pass very easily from the liquid to the curdled condition. They are often largely soluble in water, but are held in solution by a very feeble force. They never crystallise, and are generally dis- tinguished by the sluggishness of the molecules com- posing them. § 84. Dialysis. — Many insoluble colloids are per- H Digitized by VjOOQ IC 98 Hydrostatics. meable, as water is, to highly diffusive substances, but effectually resist the passage through them of other colloid substances which may be in solution. If, therefore, a solution containing a crystalloid and col- loid substance be placed in a vessel containing water from which it is separated by a septum, or membrane, formed of some insoluble colloid, the crystalloid will pass through the septum, and the other less diffusive substance will remain behind. This process of sepa- ration by diffusion through a colloid septum is termed dialysis. Experiment, — ^Take a sheet of very thin and well- sized paper having no porosity (so that if wetted on one side the other side remains dry) ; let the paper be thoroughly wetted and then laid on the surface of some water contained in a small basin of less diameter than the width of the paper. Having made a small de- pression in the paper, so as to form a cavity, place within it a mixed solution of cane sugar and gum arabic containing about 5 per cent of each. In the course of twenty-four hours the water below will be found to contain about three-fourths of the sugar, which will have passed through the paper, leaving the gum behind. What takes place during the process of dialysis may be thus explained : — A soluble crystalloid is ca- pable of separating the water from the colloid, with which it is feebly united in the septum. It thus ob- tains a liquid medium for diffusion. The soluble colloid, on the other hand, is unable to.separate the liquid from the colloidal substance of the septum, with which it is chemically combined, and is thus unable to find a way for its own passage by diffusion. Digitized by VjOOQ IC Dialysis — Osmose. 99 The process of dialysis is extensively employed for separating poisonous crystalloids from organic sub- stances with which they may be mixed ; and thus separated they yield more easily to the methods of chemical analysis. § 85. Osmose. — Intimately connected with the phenomena of diffusion is the interchange of liquids through porous diaphragms, which is known as osmose, Dutrochet ^ was the first to give his attention to this subject The endosmometer, an instrument invented by him, illustrates this process. It consists of a long tube, connected at its lower end with a membranous bag which forms a reservoir. This is filled with a solution of sugar or gum, and Fig. 45. is kept in water. After a time the liquid rises in the tube> and the level of the water falls in which the endosmometer is placed. Moreover, traces of the substance contained in the bag are found in the water out- side. This shows that an inter- change of the liquids has taken place, and that more liquid has passed inwards through the mem- brane than has escaped outwards from the reservoir. To this flowing-in of the liquid Dutrochet gave the name of endosmosis^ and he called the reverse process, by which the liquid passes out from the membranous bag, exosmosis. From numerous experiments Graham was led to ' Nouvelles Recherches sur Tendosmose et Texosmose, 1828. Digitized by VjOOQ IC ICX) Hydrostatics. infer that in order to induce osmotic action between two liquids they must each be capable of acting che- mically, but in different degrees, on the diaphragm separating them. Where porous materials are used not susceptible of chemical decomposition the osmo- tic action is very slight When either of the liquids, by means of capillarity or by its action on the septum, or possibly by both processes, has effected a passage through the diaphragm, diffusion takes place, and the liquid rises in the tube. No perfectly satisfactory ex- planation, however, of these opposing currents of fluid has as yet been given. The absorption of liquids by the spongioles of plants, and the interchange of liquids, which is con- stantly taking place in the animal body, in the pro- cesses of nutrition and secretion, are probably due to osmotic action and liquid diffusion. § 86. Holecnlar Structure of Liquids. — ^The phenomena of diffusion are interesting as affording us some insight into the molecular structure of liquid bodies. For, seeing that the molecules of one liquid are readily displaced by the molecules of another liquid into which they are diffused, it is evident that these molecules must be in a state of constant agita- tion and must be capable of continually changing their positions relatively to one another. In this way, therefore, the interchange of the molecules of two liquids in contact with each other can be explained ; and the degree of diffusibility of any soluble substance would depend on the extent of the excursion which the molecules of the solution undergo, or on the dis- Digitized by VjOOQ IC Diffusion of Liquids loi tance through which a molecule can travel before it has its direction changed by impact upon another. Thus the sluggishness of a liquid colloid would be accounted for by the very small range of the motion of its molecules \ and this explanation is supported by the fact already referred to, that these liquids are frequently highly viscous, and pass very readily into the solid state — a condition of matter in which the particles, though probably capable of revolving about an axis and of oscillating within certain limits about their mean positions, are not supposed to undergo any motion of translation whatever. d by Google t02 Pneumatics. CHAPTER VI. THE PRINCIPLES OF PNEUMATICS. XIV. General Properties of Gases. — Atmospheric Pressure, § 87. Pneumatics. — The application of the prin- ciples of dynamics to the investigation of the pheno- mena presented by gaseous bodies constitutes a branch of Hydrodynamics known as Pneumatics. § 88. Expansibility and Compressibility of Chtses. Experiments. — The characteristic properties of gaseous as distinguished from liquid bodies are shown by the following experiments : — 1. If a small quantity of gas be admitted into an empty vessel it will immediately expand, so as to oc- cupy the whole of it. 2. If a bladder containing gas be placed under the receiver of an air-pump, from which the air is gradu- ally removed, the gas contained in the skin is found to expand with the diminution of the external pres- sure. 3. If a cylindrical vessel fitted with a piston con- tain a certain quantity of gas, and the piston be pressed downwards by a weight or some other force, Digitized by VjOOQ IC A tmospheric Pressure, 1 03 the volume of the gas will be diminished ; and if the pressing force be removed the gas will expand. It thus appears that the volume, which a certain quantity of gas occupies, depends on the pressure to which it is subjected, and changes with the size of the vessel containing it § 89. The Air. — ^The earth is surrounded by a gaseous envelope reaching to a considerable height above the earth's surface. It consists of a mixture of two gases, nitrogen and oxygen, with small and variable proportions of other gases, especially of car- bonic-dioxide. By the experiments of MM. Pictet and Cailletet, previously referred to (§ 4), air and its two principal constituents have been separately reduced to the liquid state. Carbonic- dioxide has 'long since been known to be a liquefiable gas. As the majority of experiments connected with gases must necessarily be performed in air, it is very desirable, at starting, to consider some of its proper- ties, as well as the various effects due to its actioa § 90. The Air has Weight— That the air is a heavy fluid was not known till comparatively modem times. Its invisibility, and its relative position with respect to all ponderables underlying it, may have helped to conceal this important fact from the know- ledge of early observers. It appears that Aristotle,^ suspecting the truth of this fact, and wishing to verify his belief, weighed a skin first empty and then inflated with air \ and finding it to weigh the same in both cases, concluded that au: was a weightless fluid. The failure of this experiment was due to his having overlooked » B.C. 384-322. Digitized by VjOOQ IC t04 Pneumatics. Fig. 46. the fact that the skin when inflated with air cx:cupied a correspondingly larger volume, and that its weight was diminished by that of the air which it displaced, which was exactly equal to the air which it contained. Aristotle's experiment was regarded as decisive for many centuries ; and it was not till the time of Gali- leo ^ that the air was known to be a heavy fluid. It was reserved, how- ever, for later philosphers to see in this fact the true cause of a variety of phenomena which were previously unexplained. Otto Guericke^ is said to have devised the following experiment for showing that the air has weight : — By means of the air-pump, of which he was the inventor, he ex- hausted a glass globe, fitted with a stopcock, of its contained air. He then very carefully weighed the globe, and as soon as the scalebeam was perfectly horizontal he opened the stopcock. The air immediately rushed in and the globe descended. From this he very justly inferred that the additional weight which had to be placed in the other scalepan to restore equilibrium was equal to the veight of the quantity of air admitted into the globe. § 91. Measure of Atmospheric Pressure. Torri- cellfs Experiment. — To Torricelli is due a most im- portant experiment, which not only shows us that the air has weight, but also gives us a measure of its pres- sure-intensity at any point. • Born at Pisa 1564 ; died 1642. '^ Of Magdeburg ; 1602-1686. Digitized by VjOOQ IC TorricelWs Experiment, 105 Take a glass tube about 34 inches in length, open at one end and closed at the other. Fill it carefully with mercury, and, placing the thumb over the open end, invert the tube with this end under the yiq.^i, surface of some mercury contained in a cup. On removing the thumb the mercury will be found to sink somewhat, and after a time will remain stationary in the tube, with its surface- level about 30 inches above the surface-level of the mercury in the cup. Now, since the pressure-intensity is the same at all points in the same horizontal plane of a liquid in equilibrium, the pressure-intensity at any point of the external surface of the mer- cury is equal to that along c d. But the pres- sure-intensity on c D is equal to h s, where /i is the difference of level of the mercury inside and out- side the tube, and s is the weight of a unit volume of mercury. Hence the intensity of the atmospheric pressure on the external surface of the mercury is ^ j* ; and the pressure on any area of the same size as the section of the tube is equal to the weight of a column of mercury having that sectional area for base and the difference of level of the surface-level of the liquid inside and outside the tube for height. The space above the surface of the mercury in the tube is called the Torricellian vacuum. It is really occupied by mercury vapour, the effect of which on the height of the column may generally be disre- garded. § 92. EflFect of Atmospheric Pressure. — The downward pressure exerted by the atmosphere is the Digitized by VjOOQ IC io6 Pneumatics. real cause of a number of phenomena which, at one time, were explained on the supposition that * Nature abhors a vacuum.' It was observed that whenever a fluid was forcibly removed from a vessel, air or some other fluid took its place; and this tendency to occupy a vacant space was foimd to be sufficiently strong to counteract gravity and Fig. 48. other forces. Thus if the bulb of a glass vessel, the stem of which is under water, be slowly heated, bub- bles of air will rise through the water, owing to the expansion of the air in the bulb; and if the source of heat be afterwards removed, the volume of the air will gradually diminish, and the water will rise in the stem. The same effect is produced in a tube open at both ends by ordinary suction. From observations such as these it was inferred that Nature abhors a vacuum ; and this general pro- position, which expresses, though somewhat vaguely, the results of experiments made, within certain limits and under particular conditions, was sup- posed for many centuries to constitute an undeniable law of Nature. Galileo was the first to show that this so-called law was not universally true. He found that water would not rise in an empty tube to a greater height than about 2,Z feet, and he very rightly concluded that the force supporting this column of water was the pressure of the atmosphere. Nature's abhorrence of a vacuum was thus proved to have a practical limit Galileo does not, however," appear to have recognised all the consequences of his own Digitized by VjOOQ IC Barometers, 107 discovery. It was left to Pascal, by an extension of Torricelli's experiment, to give to this important prin- ciple its full development. By performing Torricelli's experiment at different elevations above the sea-level, Pascal successfully disproved the old theory, which he showed was the result of insufficient and circumscribed observations ; and by clearly establishing the fact that the column of liquid supported varied with the height of the atmosphere above it, he proved conclusively that the atmospheric pressure was the true cause of all the phenomena in question. § 93. Barometers. — ^A barometer in its simplest form consists of a tube such as that used in Torricelli's ex- periment, and containing a column of liquid supported by the atmospheric pressure which it serves to meiasure. The height of the column is the difference in level between the surface of the liquid in the tube and in the cup. This height varies with the liquid employed and the conditions of the atmosphere. fig. 49. The great specific gravity of mercury renders that liquid best adapted for the construction of a barometer, as the height of the column is corre- spondingly small. With a mer- cury barometer the average height is 76 cm. A form of instrument very com- monly in use is that known as the siphon barometer. It consists of a bent tube open at one end and closed at the other. Mercury having been introduced into the tube in sufficient quantity to Digitized by VjOOQ IC lo$ Pneumatics, fill the longer leg, the tube is placed in a vertical posi- tion when the mercury assumes a position of equili- brium, as shown in fig. 49. Its principle is the same as that of the instrument already described. The pressure-intensity at d, due to the column of mercury e d, is equal to that at c, due to the column of air above it In this form of instrument the cup containing mer- cury is not required; the difference of level of the sur- faces of the mercury in the two branches of the tube measures the pressure due to the atmosphere. § 94. Barometric Corrections. — Absolute pressure per unit area. — ^When the barometer is used for deter- mining differences of atmospheric pressure at different places, or at the same place at different times, certain corrections must be made in the observed height of the column, of which the following are the principal: — I. Correction for Temperature, — A column of mercury that measures 76 cm. at 0° C. is found to have increased in length when measured at some higher temperature, in consequence of the expansion of the liquid under the influence of heat. It is neces- sary, therefore, in comparing barometric heights at dif- ferent places and at different temperatures to calculate what the height of the column at each place would have been if the temperature had been the same; and it is usual to reduce the observed height of the column in each case to the height of a column that would produce the same pressure at 0° C. In order to ob- tain accurate results, a further correction must be made for the expansion of the scale on which the measure- ments are marked. Digitized by VjOOQ IC Barometric Corrections. 109 2. Correction for Capillarity. — ^When the mercury IS contained in a narrow tube, the internal diameter of which is less than three-quarters of an inch, the column is, as we have seen, sensibly depressed in consequence of the tension of the convex surface of the mercury. In this case the barometric reading, as reckoned from the top of the convex meniscus to the surface-level of the mercury in contact with the air, indicates a pres- sure a little less than that of the atmosphere, and the necessary correction must be added. The error due to capillarity is only observable in narrow tubes, and is slightly different when the mercury is rising from what it is when the mercury is falling in the tube. 3. Correction for Difference of Sea-level, — Owing to the compressibility of air its density at places near the sea-level is greater than at places higher up. The exact law according to which the density of the air decreases as we ascend will be considered later on ; but even if the density of the atmosphere were uni- form, the pressure-intensity would be greater at places of low than of high elevation. In fact, if a barometer is carried up a mountain it indicates a continuously decreasing atmospheric pressure ; and, from the dif- ference in height of the barometric column at two places, the difference of the elevation of the two sta- tions above the sea-level can be determined. But the height of the barometric column is liable to frequent fluctuations, owing to occasional and acci- dental causes, the effect of which it is often required to ascertain. If, with this object, observations of the barometer are made at two stations at different heights, an addition must be made to the observed reading at Digitized by VjOOQ IC I TO Pneumatics. each place, so as to reduce it to what it would be at the sea-level. Thus, if the barometric column falls, on the average, 60 mm. in being carried from the sea- level to station A, and 85 mm. to station B, 60 mm. and 85 mm. must be added to the observed reading at each of those stations. 4. Correction for the unequal value of g. — In esti- mating the pressure of the air on a given area, by the weight of a column of mercury, we obtain a result which varies with the value of ^, and affords, therefore, no uniform standard for the comparison of the atmos- pheric pressure in places situated in different latitudes. At Paris, where ^= 980*94 cm., the pressure repre- sented by a mercury column of the same height is less than at Greenwich, where ^ = 981 -i 7 cm. If h equals the height of the barometric column, and //the density of the liquid employed, the measure of the pressure of the air per unit-area x^ghd^ absolute units of force. If we put ^ = 76 cm., and//= 13*596, the density of mercury, then the absolute pressure at Greenwich per square centimetre is 98117 X76X i3'596=ioi38x 10^ cos units of force. It would be practically convenient to take the round number 10^ units of force per square centimetre as the standard of atmospheric pressure. This pressure would be represented at Greenwich by 74*964 cm., 01 29*514 inches of mercury. The standard atmospheric pressure is commonly called * an atmosphere ' ; and a pressure twice as great is known as ' two atmospheres,' and so on. Thus, a pressure of 10^ units of force would be a pressure of ten atmospheres. Digitized by VjOOQ IC The Siphon, 1 1 1 §95. Examplei.— (I.) Requiredthe height of a water baro- meter when the mercury barometer stands at 30 inches, the den- sity of mercury being 13*596. Let h be height of water barometer, A' „ ,, mercury barometer, and let d and d' be densities of water and mercury respectively. Then, atmospheric pressure per unit area ^ghd^gh'd! and .', .4 = 30 X 13-596 = 34 ft. nearly, since ^= i and g is constant. (2.) If a mercury barometer standing at 76 cm. be im- mersed 4 metres below the surface of a lake, find the height of the column. A pressure of 4 metres of water is equivalent to a pressure of 4 -^ 13*596 metres of mercury, t.^., 29*4 cm. nearly. Hence the increase of pressure will be denoted by a rise of 29*4 cm., or the height of barometric column will be 105 '4 cm. (3.) What is the absolute pressure, in units of force, due to a height of 100 metres of sea-water of density i '027, g being 981? The pressure equals 981 x 1*027 x 100 x 100= 1*0075 ^ 'o' units of force per sq. cm., and is equal to about 10 atmo- spheres. J^ § 96. The Siphon. — This is an instrument the action of which depends on the atmospheric pressure. It is used for drawing off liquid from one vessel to another, and consists of a bent tube with arras of unequal length. The siphon must be first filled with the liquid to be drawn off, and the shorter arm being temporarily closed, and then plunged beneath the liquid, a con- tinuous flow will take place. The action is thus explained : — Digitized by VjOOQ IC TI2 Pneumatics, Fio, Consider era vertical element of fluid occupying the section of the tube at its highest part. Then, if a d be a horizontal plane continuous with the surface of the liquid, the pressure on the side A c of the element c r is ^— c E, where ^is the height of a column of the same liquid, corresponding to the pressure-intensity of the at- mosphere; and the pressure on the other side is Jf—c f. Hence, as c f is greater than c e, there is a resultant force acting from right to left, and causing the liquid to flow along c d. In this way a continuous flow is maintained. It is of course understood that c e is not greater than H', for if c e were greater than If^ the liquid would commence to flow back along the limb c A. Whatever cause may diminish If diminishes the maximum height over which the liquid can be carried. Should the surface-level of the liquid in the vessel fall below b the direction of the resultant force would be changed, and the flow would consequently cease. The resultant force acting on element c r is the pressure represented by the column of liquid ^— CE — (^— Cf) = CF — CE = EF ; and, hence, the velocity of efilux is ^/2g ef, neglecting friction, and supposing the pressure at b to equal the pressure at a. d by Google A tmospheric Pressure, 1 1 3 ■' ^ ^ Exercises. IX. 1. If in ascending a mountain the barometer falls from 76 cm. to 51 cm. find the decrease in pressure on an area of one square metre. 2. A mercury barometer stands at 29*5 in., and the specific gravity of mercury is 13*6 : find the specific gravity of oil, if a column 36 ft. 6 in. in height can be supported by the at- mospheric pressure alone. 3. When the ordinary mercury barometer stands at 30 in. find the whole atmospheric pressure on a surface the area of which is 10 square feet. 4. At what depth below the surface of a lake will the baro- nieter indicate a pressure of 50 in. , when the pressure of the atmosphere is 30 in. P**^ * 5. In a siphon barometer of imiform bore the level of the mercury in the open end falls through 4 mm. : what change of pressure does this indicate ? i 6. If the sectional areas of the open and closed branches of a siphon barometer are as 4 to i, through what distance will the mercury move in the closed branch, if the mercury in the ordinary barometer rises one inch ? 7. If the specific gravity of air is 0*0013 when the baro- meter stands at 76 cm., find its sp. gr. when the barometer stands at 58 cm. 8. Supposing the average barometric height to be 30 in., the sp. gr. of mercury 13-6, and of air 0*0013, ^"^^ the height of the atmosphere, supposing the density to be uniform throughout. 9. A barometer is observeii to fall ^ of an inch when car- ried up 88 feet of vertical height : how much wovdd it fall if taken 1 10 wds up a hill rising i in 3^? 10. If the specific giravify of niercury is 13*6, what ought to be the length of a water barometer inclined to the horizon at an angle of 60^, the mercury barometer standing at 30 ins. ? 1 1. If the height of the column in an ordinary barometer is h^ and the tube is inclined through an angle of 30® from the ver- tical, what will be the barometric reading? Digitized by VjOOQ IC 1 14 Pneumatics, 12. A small bead of glass floats in the mercury of an ordinary barometer : does it affect the barometric reading ? ^ 13. Find the greatest height over which water can be car- ried by a siphon at the top of a mountain where the mercury barometer stands at 50 cm. 14. Over what height can a liquid whose sp. gr. is j be carried by a siphon, when the height of the mercurial barometer is hf and sp. gr. of mercury is /? 15. A cylindrical body 40 inches high floats in water at the ordinary atmospheric pressure with 10 inches of its height im- mersed, the sp. gr. of air being '0013. The body, with the vessel in which it floats, is then placed under a bell in which the atmospheric pressure is ten times as great : what part of the body will now be immersed ? 16. If the atmosphere press with 15 lbs. on every square inch, and the weight of a cubic foot of water be 62 J lbs., find the total pressure on a rectangular plane surface, placed verti- cally with its upper edge a foot deep in water, and its lower edge 3 ft. deep, the rectangular surface being 6 ft. long and 2 ft. wide. 17. Find an equation for determining the internal radius of a globe of thickness / and specific gravity f, which will just float in air of specific gravity j, when filled with gas the density of which as compared with air is d. 18. Taking the pressure of the atmosphere at 15 lbs. per square inch, the height of the salt-water barometer at 30 feet, calculate the pressure-intensity at 50 fathoms depth in tons per square foot. If this pressure acted for a second on a square yard of the stem of a vessel weighing 300 tons, what velocity would it communicate, there being no resistance to the motion ? XV. Relation betiueen the pressure and volume of a gas^ the temperature remaining constant. — Boyl^s Law, We have seen that the characteristic quality of a gas is its expansibility and compressibility. We come # Digitized by VjOOQIC Boyle's Law, IIS FiG» SI, now to consider the quantitive relation that exists be- tween change of volume and change of pressure, the temperature remaining constant. This relation can be inferred from a series of experiments, of which the following are instances : — § 97. Experiments. — For pressures greater than the atmospheric pressi/re, — i. A simple form of ap- paratus for these experiments consists of a uniform bent tube, having one branch open at the top and considerably longer than the other, which is closed. The shorter branch is some- times fitted with a screw-cap so that the pressure of the enclosed gas may be more easily regulated. The tube itself is gra- duated, or a scale is fixed to each branch. If the cap, which should fit au:-tight, be first unscrewed, and some mercury poured into the tube, the mercury will rise to the same level in both branches. If now the cap be screwed on again some air will be enclosed under a pressure equal to that of the external atmosphere. Call this pressure 76 cm., and suppose the mercury to stand at o in each branch. Pour more mercury into the tube, and the level of the mercury will be lower in the shorter than in the longer branch. The volume of the air confined in the shorter branch is now diminished, and its pressure is increased by that of the column c d. Suppose the air in the tube to have originally occupied 20 divisions of the tube, or 20 cm., and that mercury has been poured into the tube to a height of Digitized by VjOOQ IC o ti6 Pneumatics, 23 cm. above its former level in the longer branch, then the mercury in the shorter branch will stand at 4 cm. above o, and, consequently, the enclosed air will occupy only 16 cm. The increase of pressure, due to the difference of level, is therefore 23 — 4 = 19 cm. Now, if we examine these numbers we shall find 20 : id:: 19+76 : 76 i^e. ::95:76, or, 20x76= 16x95, which shows that the variation in the volume is in* versely as the variation in the pressure ; or, what is the same thing, that the product of the volume into the pressure is constant. V H h H-h H-h-^n-g-P PV 150 34-1 7*3 26-8 1027 1540 120 624 lO'O 52-4 1283 1539 10 -o 90-I II-8 78-3 154-2 1542 V io6-o 127 93-3 169*2 1539 8-3 I2I-3 I3'3 loS-o 1839 1526 7*4 I47-0 14-1 132*9 208-8 1545 67 i67'5 148 1527 228-6 1547 6-4 1802 150 165-2 241-1 1537 Mean value . . 1539-3 F«= volume of air In cu b. cents. ; H^ height in c ents. of merc\iry in open lira b ; h — height in closed lim b; 759 a» height of baromet er. On comparing the results of a number of actual experiments, the value of the product of the volume into the pressure will be found to vary slightly, the extent of the variation depending on the degree of Digitized by VjOOQ IC Boyl^s Law, \\^ accuracy with which the experiments are performed. In the above table are found the results of a few experiments in which the measurements were roughly made, the fractional parts of a centimetre being esti- mated by eye. They serve, however, to show how nearly constant is the value of P V. 2. For pressures less than the atmospheric pressure. To determine the relation between the change in the volume and pressure of a gas, when the gas expands, we take a barometer tube fitted with a screw-cap, and having opened the tube, partly im- merse it in a vessel containing mercury. We then close the tube, by means of the screw, and thereby enclose a certain quantity of air at the p^^ ordinary atmospheric pressure. 'J*he tube is now raised vertically upwards, and the en- closed air expands, its pressure being di- minished by that due to the difference of surface-level of the mercury in the tube and in the vessel. Thus, if ^ ^ = Fbe the volume occupied by the air when the mercury stands at the same level in the tube and in the vessel — that is, under the ordinary atmos- pheric pressure H^ and if a //= V be the volume occupied at the pressure H-cd^=^P\ then, on reference to the scale, it will be found that ah : ad:: H—cd: H, or, VH^ rP'. § 98. Statement of Boyle's* law.— From experi- * Robert Boyle born at Lismore in Ireland, 1626 ; died 1691. Digitized by VjOOQ IC nen- ' Il8 Pneumatics, ments such as these performed with different gases, and at different temperatures, a law has been estab- lished, which we shall see is not accurately true for any gas, but approximates very nearly to the truth for those gases which are not easily reducible to the liquid state. The law is known as Boyle's law or Mar- riotte's law, as these two philosophers are said to have arrived independently and by similar experiments at the same result. It may be enunciated thus : — The volume of a gds varies inversely with its pres- sure, when the temperature remains constant. Thus, if the volume V change to V\ while the pressure changes from/ to/', we have : — V : v'::p' \p) or Vp=-vy, As the mass of the gas remains the same in the experiments already described, it follows that the density of the gas must increase as the volume diminishes, and vice versL Hence Boyle's law may be stated thus : — The pressure of a gas is proportional to its density, tJu temperature remaining constant. For a perfect gas, therefore, we have the following relations : — V' p d ' Another statement of this law, due to Professor Rankine, places the law in a very clear light : — * If we take a closed and exhausted vessel and introduce into it one grain of air, this air will, as we know, exert a certain pressure on every square inch of the surface of the vessel. If we now introduce a Digitized by VjOOQ IC Dalton's L aw, 1 1 9 second grain of air then this second grain will exert exactly the same pressure on the sides of the vessel that it would have exerted if the first grain had not been there before it. Hence we may state, as the property of a perfect gas, that any portion of it exerts the same pressure against the sides of a vessel as if the other portions had not been there.' * § 99. Dalton's Law.^ — This is an extension of Boyle's law for a mixture of different gases. If several different gases, which do not act chemically on one another, are placed in a vessel, the pressure on the sides of the vessel is the sum of the pressures due to the different gases. Thus, suppose each of the gases, if in the vessel by itself, to exert pressures the intensi- ties of which are/i, ^2, /a, &c., respectively, the inten- sity of the whole pressure exerted by the mixture is P\ -^ Pz 4-/3 + &c. Rankine's statement of Boyle's law shows this to be the case for different parts of the same gas ; hence Boyle's law may be considered as a particular case of Dalton's,^ which may be thus stated : — When a mixture of several gases ^ at the same temperature^ is contained in a vessel, each produces the same pressure c^ if the others were not present, § 100. Examples.— (I.) In a bent tube, open at one end and closed at the other, the mercury stands at the same level in both branches, and the contained air occupies 30 cm. at the normal atmospheric pressure — viz. 76 cm. If the section of the tube is 10 sq. cm., what volume of mercury must be poured into the tube to compress the air to 20 cm. ? If / be the pressure of the air when occup3nng 20 cm., wc C. Maxwell, Theory of Heat, pp. 27-8. « 1801. ■ Bom in Cumberland 1766 ; died 1844. Digitized by VjOOQ IC I20 Pneumatics, have by Boyle's law 20 x /« 30 x 76, .'.;?= 114, of which 76 is due to atmospheric pressure. Hence the difference of level in the two branches equals 38 cm. ; and as the mercury has risen 10 cm. in the closed branch the quantity of mercury introduced equals (20 + 38) ID = 580 cubic cm. (2.) The mercury in an ordinary barometer stands at 30 in., and the sectional area of the tube is one square inch. A cubic inch of air is admitted through the mercury into the vacuum above, and depresses the column through 4 inches : find the size of the vacuum. • ^ When the cubic inch of air is admitted into the vacuum it first fills the vacuum, and then with its diminished elasticity, consequent on its expansion, it depresses the mercury until the pressure of the column of mercury, together with the elasticity of the air, equals the atmospheric pressure. Suppose the vacuum to measure x inches ; then the air which under a pressure of 30 inches occupied i inch is found to occupy ;c + 4 inches under a pressure of 30— (30— 4) = 4 in. Hence, by Boyle's law, 4 x (^ + 4) = 30, or ;c=3'5 in. § loi. Graphic representation of Boyle's Law. — Take o x, o^, two straight lines at right angles to each *'^G. 53. other, and along ox mark off o a, to repre- sent the number of units {V) of volume which a certain quan- _n tity of gas occupies at a -^ pressure/. Then, if we mark off o b, o c. . . to represent 2 ^ 3 F .... we know by Boyle's law that the pressures corresponding to these volumes will be ^, ^ .... Hence, if a p be drawn vertically from 23* A to represent /, then' the lines b q = — , c r = Digitized by VjOOQ IC \ K Graphic Representation of Boyle's Law. 1 2 1 — , &c. will represent respectively ^^ ?-, &c., and the 3 . ^. ^ curve drawn through these points will serve as a gra- phic representation of Boyle's law. Now, it is evident that no amount of compression can reduce the volume of a gas to zero, and no amount of expansion can wholly destroy the pres- sure which a perfect gas exerts ; it follows, therefore, that the curve will approach on either side nearer and nearer to the lines o x and oy, but will never meet them. Since, also, Vp = V'p' = V"p", we see that the product of the volume and the pressure remains con- stant, whilst the separate factors vary, and therefore the curve p q r is such that if p be any point in it, the area of the rectangle o a p a is constant for all positions of p. Now, if X represent any volume measured along ox, and J' the corresponding pressure measured ver- tically upwards, we have xy ^=z o. constant, and the curve corresponding to this equation is known as the rectangular hyperbola. § 102. Graphic representation of the work done in changing the volume of a gas from Fto V'^ the temperature remaining constant. Let a volume V of gas be en- fig. 54. closed in the cylindrical vessel d b a c A yy c by the piston m n, and let the pis- ton be moved through m m' so that ^ ^ w i> the volume becomes V, Then if p be the pressure of the gas, when the volume is V^ the work done in changing the volume to V would be / x m m', Digitized by VjOOQ IC 122 Pneumatics. supposing / to remain the same throughout. But this is not the case, for the pressure increases as the volume diminishes. If, therefore,/' is the pressure of the gas when its volume becomes F', the work done is equal to some quantity the value of which Ues between / x m m' and />' x m m . Now, suppose o a (fig. 55) represent the volume V^ and o a' the volume F', a p, a' p' the correspond- F»G. 55. ing pressures ; then, if we draw the lines p q, p' q', parallel to o A, the work done is repre- sented by the area of a figure which is greater than the rect- angle p Q a' A, and less than the rectangle p' a' a q'. Now, by reasoning similar to that employed for finding the space described when a body moves with an increasing velocity, it can be shown that the area of the figure I p' a' A represents the work done. § 103. Limits of Boyle's Law. — By experiments conducted by Regnault ^ and by Despretz* it has been observed that Boyle's law is not perfectly true for any actual gas. For air and all gases that do not readily liquefy under pressure the law is found to be a very close approximation to the results of experiments ; but for easily liquefiable gases, such as carbonic di- oxide and ammonia, the volume is found to decrease more rapidly than the pressure increases ; and the divergence of the law from the true results is greater as the gases approach their point of liquefaction. » 1827, 1847. * 1827. Digitized by VjOOQ IC Limits of Boyle's Law. 123 Thus carbonic-dioxide under a pressure of twenty atmospheres is found to occupy only four-fifths of the volume given by the law. Regnault devised very careful experiments for showing the exact relation between the pressure and volume of different gases and the extent of the diver- gence of this relation from Boyle's law. He arrived at the following result : that whereas, according to Boyle's law, Vj> = P/, or -/, -1 = 0, the ex- Vp Vp' pression -^J--, — i is a quantity having a small posi- tive value for all gases except hydrogen, and increases gradually with the pressure. If the distances o a, o b, o c, measured along the horizontal line o x represent pressures of one, two, Fig. 56. and three atmospheres, and if the vertical lines a a, B b, c c represent the values of ^^ — i for these pressures, the curve formed by joining o « ^ ^ . . . is a graphic representation of the divergence of the results of direct experiment from Boyle's law. Seeing that f^ — i is a positive quantity, Vp Digitized by VjOOQ IC 124 Pneumatics. must be greater than V p', and therefore the value of V must be less than it would be if Vp = V' p\ in accordance with Boyle's law. Hence it appears that gases are more compressible than is in accordance with Boyle's law. In this respect, however, hydrogen differs from other gases, and this exception shows that the law is more complicated than it would seem to be even with the extension above given. It has, however, been proved that the true relation of the volume of a gas to its pressure depends to some extent on the tempera- ture. The curve exhibited in fig. 56 is not the same for all temperatures. For nearly all gases the value of V V the expression -j^ril, — i contmually decreases as the V p temperature rises, and we are 'thus led to expect that if the temperature were sufficiently high this quantity would pass through zero and become nega- tive ; or the curve, after coinciding with the line o Xy would reappear on the other side. It should seem, tlierefore, that at a certain temperature varying with each gas Boyle's law is strictly accurate, and that for higher temperatures the law of the divergence is changed, so that the density increases less rapidly than the pressure. Now, as hydrogen which has already been reduced to the solid state is commonly supposed to be the vapour of a metal, it is relatively at a very high degree of rarefaction, and this fact may be the reason why — ^ - i has a negative value for this gas. § 104. Belative Densities of the Air at di£Eerent Digitized by VjOOQ IC Variation in Density of Atmosphere, 125 Heights. — ^We are now in a position to determine the law according to which the density of the air changes as we ascend from the level of the sea. The decrease in the density of air is owing to its compressibility; but even if the air were as incompressible as water, and the atmosphere were homogeneous throughout, the baro- metric column would be found to fall, in rising from places of lower to places of higher elevation, in conse- quence of the diminution in the height of the column of air. In order to obtain an approximate relation between the densities of the air at two different heights we must neglect the accidental differences of pressure caused by differences of temperature and moisture, and by the altered value of the force of gravitation. Take a vertical column of the atmosphere and sup- pose it divided by horizontal planes Fig. 57* into a number of strata so thin that the density for each layer may be con- sidered uniform and equal to that at its lower surface. Let the height of the column be z^ and let n be the number of _ strata, so that the thickness of each layer is - n let the section of the column be the unit of area. Let d and p be the density and pressure of the atmosphere at the surface of the earth, and let //j, d<i^ d^ • • 'iP\ip2iPz' • • • be the densities and pressures at the successive levels. Then, since the difference between the pressures at the upper and lower surface of a layer must be equal to the weight of that layer of air, p-px = weight of lowest layer = dg^ (§ 17): d by Google 126 Pneumatics And since by Boyle's law the pre density, we have -^ = -^^ -^ = a dx d^ k {d -- dx)^a whence ^/i = (i — S. \ k similarly ^, = fi^^t \ kfi />., the ratio of the densities for t^ is constant, since g, k^ z and n are fore the quantities d^ d^^ d^, d^ . . cal progression, the heights of tl arithmetical progression. If, therefore, dn be the density d \ kn) % 105. To find the difference stations by means of a baromet temperature and force of gravitj Let H and h represent the ba at the two stations, the vertical dist being z^ ThenIf:A::d: a If V k y Google Difference of Height and Pressure. 1 27 where d\% the density of the air at the lower, and d^ the density at the upper station. Now, \in increase without limit, it is proved in works on Algebra, that the value of the right-hand side of this equation becomes e Hence ^ = ^ > and \og.S-i ] == f * ^ Or,0=^(log.e^-log.e>^) Exercises. X. 1. A gas occupies 100 litres when the barometer stands at 76 cm, : find the increase in the volume of the gas, if the pressure becomes 73 cm. 2. A tube 2 feet long is filled with water and inverted in a vessel of water, with its open end below the surface. Air at a pressure of 30 ins. is then admitted into the tube till the level of water in the tube is the same as that outside, and the air occupies 12 inches. The tube is now raised till the air occupies 15 inches : find the pressure of the contained air. 3. Into the vacuum above a common barometer, which stands at 30 inches, 2 cubic inches of air are admitted, which depresses the mercury 6 inches : if the section of the tube is one square inch, find the size of the vacuum. 4. A vessel of 3 cubic feet capacity containing air at two atmospheres' pressure is put into communication with a vessel of 18 cubic feet capacity containing air at \ of the atmospheric pressure : what is now the pressure of the air in the two vessels ? 5. A horizontal cylinder containing air is fitted with a piston which is 10 inches from the closed end when the pressure is 15 lbs. on the square inch. If the area of the piston is 8 square d by Google 1 28 Pneumatics, inches, find the force that must be exerted to hold the piston at a distance of 12 inches from the closed end. 6. If the water barometer stand at 33 feet, to what depth must a small cylindrical vessel be sunk to reduce the volume of the contained air to one-third of its original volimie, the height of the vessel itself being neglected ? 7. A cylinder contains air at the ordinary atmospheric pres- sure, and is fitted with a piston (area 10 square inches), which is I ft. from the closed end. If the cylinder be set vertically with its open end upwards, how far will the piston descend, the atmospheric pressure being 15 lbs. on the square inch, and the piston weighing 10 lbs. ? 8. Suppose the cylinder is held with the open end down- wards, how far will the piston fall ? 9. Find what weight must be hung to the piston in the last question to draw it down 2 inches from its original position. 10. The air contained in a cubical vessel the edge of which is one foot, is compressed info a cubical vessel, the edge of which is one inch : compare the pressures on the side of each vessel. 11. Forty c.c. of air are enclosed in a tube over mercury, the height of the mercury in the tube above the level in the vessel outside being 50 cm. (r ^«5o, fig. 52). The tube is de- pressed until cd='}fi cm. What is now the volume of the air, the height of the barometer being 76 cm. ? 12. A bent tube (fig. 51) has a uniform section of I square inch and is graduated in inches ; 6 cubic inches of air are en- closed in the shorter branch, when the mercury is at the same level in both branches. What volume of mercury must be poured into the longer branch in order to compress the air into 2 inches ? The barometer stands at 30 ins. 13. The air enclosed in the shorter branch of a similar bent tube occupies 11*3 cc, and the difference of level in the two branches is 60*2 cm. If the barometer stands at 75*9 cm., find what volume the air would occupy under the atmospheric pressure only. d by Google Experiments oft Diffusion of Gases, 129 XVI. Diffusion of Gases. § 106. DiflFtision of Oases. Experiments. — We have seen that if two layers of different liquids are in contact with each other, or are separated by a porous diaphragm, a mixture of the liquids takes place, the one diffusing into the other. Now, the same pheno- menon is observable with gases ; but the laws of gaseous diffusion are less complex than those of liquid diffusion, in consequence, probably, of the greater structural simplicity of gaseous bodies. Fill two jars with two different gases — for example^ with chlorine and hydrogen — and let the jars be connected by a long tube, that containing the hydrogen or lighter gas being placed uppermost In a few hours the chlorine will find its way into the upper jar, as may be seen by its green colour; and the hydrogen will take its place. Each of the jars ^^^ ^ will be found to contain the same pro- portion of the two gases, and the gases will remain permanently mixed. This intermixture .takes place between any gases or vapours which do not act chemically on one another. If a vessel containing nitrogen be covered with a porous diaphragm of some colloid substance, and be placed under a bell-jar containing hydrogen, /lIU^ diffusion will take place, and after a time the membrane will have become convex, showing that the hydrogen has been passing inwards more Digitized by VjOOQ IC 1 30 Pfieumatics, rapidly than the nitrogen has passed outwards. If the position of the gases be reversed the contrary result will be found. If the diaphragm be moist, and one of the gases is soluble, its rate of diffusion is very much increased. Thus, if a moist thin bladder be dis- tended with air and placed in a bell-jar containing carbonic dioxide, this gas, owing to its solubility, passes much more rapidly into the bladder than the air escapes from it, and very frequently breaks it, though the rate of diffusion of carbonic dioxide into air is really less than that of air into the gas. § 107. Rate of DiflFasion. — If we take a long gra- duated tube, open at both ends, and close one end by a plug of porous clay, and immerse the open end in a vessel containing mercury, or water coloured, for the sake of greater distinctness, the level of the liquid in the tube and in the vessel will be the same, showing that the gas must be entering the tube through the porous plug at the same rate as It escapes mto the outer atmosphere. If now we bring an inverted beaker filled with coal-gas over the closed end of the tube, we find that there is a bubbling of gas through the liquid, clearly showing that the coal-gas is entering the tube more rapidly than the air is escaping from it. As soon as we withdraw the beaker the liquid rises in the tube, the pressure of the 7 mixed gas in the tube being less than that of the atmosphere out- side ; and if we replace the beaker Digitized by VjOOQ IC Graham's Law. 1 3 1 the bubbling of the gas through the liquid recom- mences. This experiment enables us to observe the difference only between the amount of coal-gas that goes in and of air that escapes from the tube. We can vary this experiment by filling the tube with other gases than air. Suppose the tube to be first filled with hydrogen and then inverted in the liquid, and held in such a position that the level of the liquid within and outside the tube is the same, the beaker being altogether removed. After a time the liquid will be found to rise in the tube, showing that the pressure within the tube has become diminished, and that the hydrogen must be passing from the tube into the external air more rapidly than the air is entering the tube. It is easy to see that, if the tube be graduated, careful experiments will show the volumes of different gases that diffuse into the air in the same time. If the beaker be filled with a different gas, and held over the tube, the rates of diffusion of different gases into each other can be ascertained. By experiments such as these, Graham established the law that the rates of diffusion of two gases into each other are in the inverse ratio of the square-roots of their densities. For instance, taking the density of air as unity ^ that of hydrogen is 0*0692 ; and the square- roots of these numbers being i and 0*2632 respectively, the law tells us that the rate of diffusion of hydrogen is to the rate of diffusion of air as I : I-^o•2632, i.e. \\i \ 37994; and actual experiment shows that whilst one measure K 2 Digitized by VjOOQ IC 132 Pneumatics, of air passes into the tube containing hydrogen, 3*83 measures of hydrogen escape into the air. This important law can be . further verified by taking a vessel consisting of two large receivers filled with different gases, and connected by a tube with a stop-cock. If now the gases be allowed to diffuse into each other for a certain period of time, the contents of the receivers can be analysed, and the proportion of the two gases in each can be quantitively deter- mined. By varying the time, and tabulating the results, the accuracy of the law may be verified for any two gases. The following table shows the results of some of Graham's experiments : — Gas Density Square root of density z Rate of diffusion VDensity Hydrogen .... Marsh-gas .... Carbonic Oxide . . Nitrogen .... Oxygen .... Nitrous oxide . . . 0*06926 0-559 0*9678 0*9713 1*1056 1*527 0*2632 0*7476 0*9837 0*9856 I 0515 1*0914 37994 1-3375 1*0165 1*0147 0*9510 0*8092 3-83 1*34 I 0149 1*0143 0*9487 082 If d and d' be the densities of two gases, and D and D' the volume of each which diffuses into the other in the same time, then, according to the law, D:D'::^A=:i ^^, I \/ d s/d OxD^d^D'^d'. § 108. Kinetic Theory of Oases. — In order to ex- plain the diffusion of gases we must suppose that the Digitized by VjOOQ IC Kinetic Theory of Gases, 133 particles of a gas axe independent of one another, that they are constantly moving in all directions, and with a very great velocity. These particles during their motion frequently impinge on one another, and the direc- tion of their motion is consequently changed. "When they come into contact with the sides of the vessel containing the gas their momentum is resisted ; and it is to this shower of particles moving with a consi- derable velocity that the pressure of a gas is supposed to be due. The notion that the particles of a gas are in rapid motion, and that it is by their impact that gases press on one another, is a very old one. It was pointed out, not long after Newton's time, by Daniel Bemouilli ^ ; Lesage and Prevost of Geneva made several applications of the theory, and it was afterwards revived in this country by Herepath. In 1848 Dr. Joule showed how the pressure of gases might be explained by the impact of their molecules, and he calculated the exact relation that exists be- tween the observed pressure of a gas and the velocity of its particles. It is, however, to Professors Clausius and Clerk-Maxwell that this theory of the molecular structure of gases owes its chief development. By a method similar to that indicated in Wor- meirs * Thermodynamics/ § 71, Joule showed that if a vessel contain hydrogen at the ordinary pressure and at 0° C, the velocity of the particles must be about 6,055 feet per second. Now, although the velocity of these particles is so considerable, the number of particles occupying a given volume, say a cubic inch, is so enor- * Born at Groningen 1700 ; died 1782. Digitized by VjOOQ IC 134 Pneumatics. mous that the particles move through a very small space, and are unable to travel from side to side of the vessel containing them, without encountering a series of successive impacts with other particles. It follows from this, that if the average velocity of the particles remains the same, the pressure exerted at any point of the vessel containing the gas depends on the num-' ber of particles that impinge, in a given time, on the element of area containing that point But this, of course, depends on the number of particles contained in the vessel, />., on the density of the gas ; for if the volume of the gas be doubled, the average velocity of the particles remaining the same, the number of particles traversing a given area will be halved. Now, the density of a gas is the ratio iDf its mass to its volume, and hence it follows that the pressure a gas exerts varies inversely with its volume, if its mass or the number of particles in a given volume remain the same; and this result is the same as that previously obtained by experiment, and known as Boyle's law. The notion that a gas consists of a series of parti- cles flying about in all directions is the basis of what is called the kinetic theory of gases. This theory, besides explaining the phenomena of diffusion and Boyle's law, accounts for many other facts connected with the action of gases at different temperatures; but this subject is beyond the range of the present volume. d by Google Diving Bell. 135 CHAPTER VII. PNEUMATIC INSTRUMENTS. XVII. Diving Bell, Pressure Gauges, The two principles which we have now established — viz., that the air is a heavy elastic fluid, and that the density of a gas varies with its pressure — serve to ex- plain most phenomena exhibited by gases at constant temperature, and enable us to understand the action of a great variety of pneumatic instruments. § 109. The Diving Bell.— If we take an ordinary glass tumbler and immerse it vertically, with its mouth downwards, in a tub of water, we shall find that the enclosed air will prevent the water from rising in the inside except to a very small height, which will vary with the depth to which the tumbler is immersed. The action of the diving-bell is similar. It consists of a hollow vessel, nearly cylindrical in form, and open at its lower end. When lowered vertically into water the enclosed air is compressed by the weight of water above it, and the water rises in the bell to a height which increases with the depth of the bell from the surface of the water. The bell is let down by a chain; and in order that men may be able to work in the inside, air is introduced by means of a pump through an opening in the top, and the pressure of the air prevents the water from rising in the bell. Digitized by VjOOQ IC 136 Pnemnatics, § no. Problems on the Diving Bell.— (i.) To find the h'iight to which water rises in the bell, at a given depth, when no additional air is introduced : — Let B c = 0, be the depth of the top of the bell below the surface of the water. Let b be the height of the bell, and H the atmospheric pressure at the surface of the water measured by a water baro- meter. Then, if c a = x^ the part of the bell occupied by the air at the depths, and if we suppose the bell to be of uniform area inside, the pressure on the air within the bell is equivalent to the weight of a column of water the height of which is Zr+ BA = ^+^ + ^. Hence, by Boyle's law :— X H FIO. 6<>. Ma^^ WeM 3£r:^ 1^ i^i^ ^^^ A £=£^ rrrr.. . 1^^ '■fzzi b If+z+x or, x^ + X (If + z) = Hb a quadratic equation, the positive solution of which gives the height required. (2.) To find the volume of air at the ordinary atmospheric pressure that must be introduced into the bell at a given depth, to prevent any water from entering: — Let z be the depth of the top of the bell, and b the height of the bell, as before. Then, if V be the volume of the air in the bell at the normal pressure H, and F' the volume which the compressed air in the bell at depth z + b would occupy at pressure H, it follows Digitized by VjOOQ IC Manometers, 137 from Boyle's law that V H i\e. the volume of air introduced, at the ordinary atmospheric pressure, is T, . V. H § III. Manometers, or Pressure-Chkuges. — Mano- meters are instruments for measuring the elastic force of a gas contained in a closed vessel. fig. 6x. The simplest form of manometer con- sists of a long narrow open tube which dips into a strong box containing mercury. The gas, the pressure of which is to be measured, is admitted into the box through an opening, a, and if the elasticity of the gas is equal to that of the air the level of ^ the mercury in the tube will be the same | as that in the box. If, however, the L elasticity of the gas is greater, the mercury will be forced up the tube, and the excess of the pressure of the gas over that of the atmosphere can be measured by the height of the mercury in the tube. This form of manometer cannot be used for very great pressures, as the length of the tube renders it inconvenient. Thus, for a pressure of two atmo- spheres the tube must be 30 inches, and the length of the tube must be increased 30 inches for every addi- tional pressure of one atmosphere that is to be measured. d by Google 138 Pneumatics, § 112. Compressed Air Manometer. — For mea- suring greater pressures this form of instrument is better suited. ^^^' ^ _ It consists of a bent tube, one branch of which is closed and contains air at the ordinary atmo- spheric pressure. This air is shut off from the other branch by some mercury which occupies the lower part of the tube. The open branch communicates with the vessel containing the gas, the pressure of which is to be measured. The closed branch of the tube is fiunished with a scale. If the mercury stands at the same level in both branches of the tube the pressure of the gas will equal that of the atmosphere. But if a gas or vapour of greater pressure be admitted through g the level of the mercury will fall in d and rise in c above the ori- ginal level, A B. The measure of the pressure of the gas in d is that of the compressed air in c, together with that indi- cated by the difference of level of the mercury in the two branches of the tube. In order to graduate the scale we must find the distance a e or rise of the mercury conesponding to a pressure of n atmospheres in d f. Let a c, the space originally occupied by the air, equal «, and let AE =^. Let H be the height of the mercury in the baro- meter at the atmospheric prebsuie, and/ the pressure of the air in c e. Digitized by VjOOQ IC Pressure-gauges. 139 Then/xcE = ^x ac, or 4 = :i5 = _^ and the pressure of the gas in df=/ + 2Ea = n If. a—x /. 2x'^^ (nH-\-2d)x + {n—i)Ha=^o-y and, by the solution of this quadratic, the value of x^ corresponding to any number of atmospheres, can be determined and the scale graduated. § 113. The Siphon Gauge.— For measuring small pressures this instrument is sometimes employed. It consists of a bent tube, open at both ends, one of which communicates with the vessel containing the gas. The liquid used is water or mercury. If the gas is admitted through b, and the liquid assumes a difference of level, p d, then the pressure of the gas equals the atmospheric pres- fig. 63. sure + the weight of the liquid in p d. If, however, the liquid rises in the other branch of the tube, then the pressure of the gas = the pressure of the atmosphere, minus that due to the difference of level. Thus, if the liquid fall through x inches in one branch, it will rise through x inches in the other ; and the difference of level will be 2 X. If, therefore, a be the sectional area of the tube and s the specific gravity of the fluid — The pressure of the gas = atmospheric pressure ± 2XS, according as the level sinks in b or a. d by Google 140 Pneumatics, Exercises. XI. 1. To what depth must the top of a diving-bell 8 ft. high be immersed under water that the air may be compressed to half its volume, the height of the water barometer {H) being equal to 34 ft.? 2. What additional volume of air at the ordinary pressure (^=34 ft.) must be admitted into a bell 8 feet high, the internal section of which is uniform and equal to 20 sq. ft., and the top 60 feet below the surface, to completely fill it ? 3. A cylindrical bell, the height of which is 6 feet, is fur- nished with a barometer that stands at 30 in., and is lowered into water till the barometer stands at 40 in. : find the depth of the top of the bell below the surface of the water ; sp. gr. of mercury = 13*6. 4. A barometer marking 30 in. is carried down in a diving- bell which is kept constantly full of air : find the depth of the top of the bell from the surface of the water, when the barome- ter marks 42 in., the height of the bell being 8 feet. 5. A gas, the pressure of which is 10 lbs. on a square inch, communicates with a siphon gauge (fig. 63), the section of which is I square inch : find the difference of level, supposing the in- strument contains mercury and the ordinary atmospheric pres- sure is 15 lbs. on the square inch. 6. If in the open manometer the distance of the level of the mercury from the top of the box is </, and a gas be admitted that depresses the mercury b within the box, find the height of the mercury in the tube above the original level, the atmospheric pressure being h, . 7. A cylindrical bell, 4 feet deep, whose interior volume is 20 cubic feet, is lowered into water until its top is 14 feet below the surface of the water, and air is forced into it until it is three-quarters full. What volume would the air occupy under the atmospheric pressure, the water barometer being at 34 feet ? 8. Find the depth to which a cylindrical diving-bell 8 feet d by Google Pumps, 141 high must be sunk in water in order that the water may rise in it 3 ft. (^= 34 ft.) 9. If the weight of a cylindrical bell is 2,000 kils. and the specific gravity of the material 8, find the tension in the chain when the bell is immersed to such a depth that the pressure of the enclosed air equals 3 atmospheres, the interior volume being I '5 cubic meters. Fig. 64. c P XVIII. Air-Pumps. § 1 14. Essential Parts of a Pnmp.— -A pump is an instrument for removing a fluid from a reservoir, or vessel containing it, by the forcible withdrawal of the atmospheric pressure. It consists essentially of (i) a barrel or cylinder^ c, through which the fluid escapes ; (2) a disc ox piston ^ p, capable of moving up and down the cylinder, into which it exactly fits, and worked by a handle or rody r, attached to it; (3) a pipe^ E, that communicates with the reservoir or vessel from which the fluid is to be removed ; and (4) valves or small apertures, with movable covers, opening one way only, v, v', which serve to admit the fluid from one part of the instrument to another, and to prevent it from returning. A valve is generally found at the end of the pipe, where it communicates with the cylinder, and very frequently in the piston itself. § 115. The Air-Pnmp. — ^This is an instrument for removing air from a closed vessel. Digitized by VjOOQ IC ^ 142 Pneumatics. Pio. 65. It consists of a receiver^ r, from which the air is to be removed ; of a cylinder or barrel, a b, furnished with a piston and valves; and of a pipe which serves to con- nect the receiver with the cylinder. The action of the pump may be thus explained: — Suppose the pis- ton at B, when the receiver, r, is full of air. As the piston is raised the valve m, which opens upwards, re- mains closed in consequence of the pressure of the external air, and the air from r rushes through the pipe E, opens the valve n, and occupies the space between the piston and the bottom of the cylin- der. When the piston is first pushed down the air in the cylinder is compressed, the valve m remaining closed j but as soon as the pressure of the air in the cylinder begins to exceed that of the air outside the valve M is opened, and the enclosed air escapes as the piston descends through it. The piston being again raised, the same process is repeated. It should be noted that it is only when the piston is raised that air is withdrawn from the receiver, and that when it descends the au: so withdrawn escapes i to the outer atmosphere. § 1 16. To determine the density of the air in the receiver after any number of strokes. Let Fand v be the volumes of the receiver and cylinder respectively. Let d be the original density d by Google The A ir-Pump, 1 43 of the air in r, when the piston is at b. Then Vd equals the mass of air in r. When the piston rises to A the mass of air in r occupies the space V-^-v ; and if di be its diminished density we have r^=(r+e/)//„or//, = y^ • ^, since the mass remains the same. When the piston descends to b a part of the air escapes, and the mass of air in r = F//i. If now d.2 represent the density of the air when its volume increases to V + v, we have Vd,={V+v) d,, or d, = {y^y d, and similarly if d^ represent the density of the air m R after three complete strokes — Vd^ = ,^{V-^v) d,, or d, = {jr^' d, and, consequently if d^ be the density of the air in the receiver after n complete or double strokes It will be seen that no amount of exhaustion can re- duce the density of the air to zero. At each stroke of the piston a fraction only of the air in the receiver is removed ; and as the remaining air occupies the whole volume of the receiver, a perfect vacuum can- not possibly be obtained. § 117. Examples.— ( I.) If the volume of the receiver is 64 cubic inches and of the cylinder 8 cubic inches, what quan- Digitized by VjOOQ IC 144 Pneumatics. tity of air would be left in the receiver after two complete strokes ? At the first double stroke 8 cubic inches of air at the original density would be removed and 56 would remain. Of these one- eighth, or 7 cubic inches, would be removed at the second stroke, and consequently 49 would be left. (2.) After four complete strokes the density of the air in the receiver is to its original density as 10,000 : 14,641 : compare the volumes of the receiver and cylinder. d 14,641 \K + z// F+v V 14,641 II § 118. Difficulty of Workiiig.— It is to be ob- served that the difficulty of working this kind of air- pump increases with the number of strokes. For, neglecting the frictional resistances^ the force required to raise the piston depends on the difference of pres- sure at ♦the upper and lower surfaces of the piston. Now, this difference increases with the rarefaction of the air in the receiver, and consequently the difficulty of working the machine increases as the exhaustion proceeds. In lowering the piston the external pressure as- sists the action, so long as the air in c is of less den- sity than that outside ; but as soon as that point has been reached an expenditure of force is necessary to overcome the inside pressure and open the valve m. Thus the difficulty of lowering the piston decreases somewhat with the number of strokes. § 119. The Double-barrelled Air-Pump. — This instrument, known as Hawksbee's air-pump, has two Digitized by VjOOQ IC The Double-barrelled A ir-pump. 145 Fig. (>t. cylinders, in each of which is a piston worked by a rack and pinion. To the centre of the toothed-wheel is fixed a handle, by moving which to and fro the pistons are made alternately to as- cend and descend. The chief advantage of this instrument over the one already described . is that a volume of air equal to that of the cylinder is removed at each single stroke of the piston, and that, consequently, the rate of exhausting the receiver is doubled. In the adjoining figure the piston a is ascending and withdrawing air from the receiver, and the piston B is descending and discharging into the outer atmo- sphere the air previously withdrawn from the receiver. The position of the valves should be carefully noted. Another advantage of this machine is that it is easier to work. The difficulty of raising one piston is compensated by the assistance which the pressure of the atmosphere affords in forcing down the other piston, and thus the difficulty of working the machine does not increase, as in the case of the single-barrelled air-pump, with the number of strokes. In pressing down the piston no force is required till the air beneath the piston has been compressed to the density of the air outside ; and this occurs nearer and nearer to the bottom of the cylinder as the degree of exhaustion increases. Consequently, the instrument is worked somewhat more easily as exhaustion proceeds. Digitized by VjOOQ IC 146 Pnmmatics. § 120. Tate's Air-Pump. — This instrument combines the advantage of double action with a single barrel. The barrel is generally horizontal, and communicates Fig. 67. with the receiver by a vertical opening, o. The pis- ton, c D, occupies a little less than half the length of the barrel, and consists generally of two discs rigidly connected together by the piston-rod which unites both. The principle of the action would be the same if the piston were solid and of uniform area through- out ; but the trouble <A working it would be greater. The barrel is furnished with two valves, a and b, at either end of it, opening outwards. The action may be thus explained : — Suppose the piston to be, first of all, in the posi- tion shown in fig. i, all the air in firont of it having been expelled through a by the driving of the pis- ton home. If the piston be now pulled out, as in Digitized by VjOOQ IC Tate's Air-pump, H7 fig. 2, the valve a will close, and the air in front of the piston will pass through b, which will remain open. When the piston reaches the farther end, b, as shown in fig. 3, air will escape from the receiver through o into the empty space left behind the piston. When the piston is pushed inwards this air will be expelled through A ; and on the piston reaching a, as shown in fig. I, the empty space behind it will be again occupied by the air firom the receiver. In this way, a certain volume of air, equal to about half the contents of the barrel, will be removed at each stroke of the piston ; and, as the difference of pressure at the two ends of the piston decreases with every stroke, the working of the pump becomes easier as the exhaustion pro- ceeds. § 121. Mercury Chmge. — The pressure of the air, after any number of strokes, in the receiver of an air- pump is indicated by a gauge, which is fig. 68. generally attached to the connecting- = pipe of the air-pump. In its simplest form it consists of a straight tube, open at both ends. The upper end is connected with the receiver, and the lower end dips into a cup of mercury. As the air is removed firom the receiver the pressure inside the tube is less than that on the mercury in the cup, and consequently the mercury rises "" in the tube. This instrument enables us to watch the process of exhaustion from the very first stroke. If the barometer at the time marks 30 inches, and the mercury has risen 4 inches in the tube, the pressure L 2 Digitized by VjOOQ IC 14^ Pneumatics. Fig. 69. of the air in the receiver is 30— 4 = i6 in. The necessary length of this form of gauge renders it somewhat inconvenient. The more commonly employed gauge consists of a bent tube (fig. 69), having one end closed and the other open. Each branch is about ten inches in length, and the closed end is filled with mer- cury, the weight of which is supported by the atmospheric pressure. The instrument is I gg II enclosed in a glass case, which _!lL I I communicates with the re- ceiver of the air-pump. The first few strokes do not produce any change in the gauge, but as soon as the tension of the au: in the receiver is less than the pressure due to the column of mercury in the closed end of the tube the mercury begins to fall, and the difference of level of the mercury in the two ends measures the pressure of the air in the receiver. The open end is SQmetimes bent again upon itself, as in fig. 70, and screwed into the connecting pipe. § 122. Experiments. — ^The experiments that can be performed with the air-pump are very numerous. We have found it necessary to refer to some of them, in order to prove that the air has weight (§ 90). The following additional experiment should also be per- formed : — Take a hollow cylindrical vessel and stretch a bladder over one end, and then place it on the plate of the air-pump, having carefully greased the ^dges of the glass vessel, so as to prevent the entrance Digitized by VjOOQ IC Magdeburg Hemispheres. 149 of the air. As the air from ^ under the bladder is gradually removed the bladder will be found to yield under the pressure of the external air, and will at length break. By means of the Magdeburg hemispheres, in- vented by Guericke, some idea may be formed of the magnitude of the pressure which the atmosphere exerts on a comparatively small surface. Their gene- FlG. 71. ral form is shown in fig. 71. The edges are greased and pressed together, and the enclosed air is then removed by the air-pump. In order to separate the two hemispheres by pulling them asunder, a very great effort must be made ; and this will be found to be the case in whatever position the hemispheres may be held, showing also that the atmospheric pressure acts equally in all directions. d by Google 150 Ptieiimatics, § 123. Sprengers Air-Pump. — By means of the following contrivance a small receiver can be more effectually exhausted than by any of the pumps already described. It consists of a tube, f b, open at both ends, and fitted to a funnel a, by a piece of indiarubber. The funnel contains mercury, the flow of Fig. 72- which through the tube can be regu- ^Sil lated by tightening the indiarubber ▼^ connection. The tube is consider- ably longer than a barometer tube, and has a spout in its side attached to the receiver, e, which is to be ex- hausted. The lower end of the tube, B, dips into a vessel of mercury. As soon as the mercury begins to flow ex- haustion commences. The first drops of mercury that run out close the A lower end of the tube and prevent B M. the air from entering. As each drop fiy^^^ of mercury passes the neck, c, the jr|WS|^. air in the receiver, e, expands and ^''^^^^^^ occupies the space cq. In this way the tube becomes filled with cylinders of air and mercury, which gradually escape firom the spout into the vessel of mercury. The mercury is poured again into the funnel, a, and the process is repeated till the tube is occupied by a continuous column of mercury, the height of which is equal to that of the mercury barometer. The exhaustion is now complete, and the receiver, e, corresponds to the Torricellian vacuum gf the Qrdinary barometer. Digitized by VjOOQ IC Sprengers A ir-pump, 151 § 134* ^^ Condensing S]rrmge.-— This consists of a barrel, a b, into which an air-tight piston fits, having in it a valve that opens downwards. At the bottom of the barrel is a valve that likewise opens downwards and communicates with a re- ^^°* ^3- ceiver, to which the syringe is tightly screwed. When the piston is moved down, the valve b is closed and the valve A opened by the increased pressure. As the piston returns, the pressure of the air in the receiver closes the valve a, and thus prevents the air from re-entering the barrel It is evident that the same quantity of air will enter the receiver at each stroke of the piston, if the machine works per- fectly and the piston is moved through the whole length of the barrel. Exercises. XII. 1. If the volume of the receiver of an air-pump be eight times that of the barrel, compare the density of the air after the third stroke with its original density. 2. If the receiver of an air-pump holds 90 grains of air at the ordinary pressure, and if the barrel can hold 10 grains, what will be the weight of the air in the receiver after four complete strokes ? 3. If one-third of the contents of the receiver is removed at each complete stroke, and the barometer stand at 75 cm., find the height to which the mercury will rise in the simple barometer gauge after three strokes. 4. The mercury rises in the barometer gauge through 2^ in. in two complete strokes : compare the size of the barrel with that of the receiver {h = 30). Digitized by VjOOQ IC 152 Pneumatics, 5. The brianches of a siphon barometer gauge are each 16 cm. long, and the closed branch is filled with mercury, which also occupies I cm . of the open branch : compare the den- sity of the air in the receiver with its original density, when the mercury begins to fall in the siphon-gauge (^ = 75 cm.). 6. A receiver attached to an air-pump has the volume of 100 cubic inches, while the cylinder has the volume of 10 cubic inches. What proportion of the original air will be left in the receiver after the completion of the fourth double stroke? 7. The capacity of the barrel of a condensing air-pump is 10 cubic inches, and of a copper receiver 100 cubic inches. By how much will the pressure of the air in the receiver be in- creased after 20 strokes of the piston ? 8. When the height of the barometer is 75 cm.^ the air in the receiver of an air-pump is exhausted until the mercury in the barometer-gauge (fig. 68) attached to it rises from o to 36 cm. By how much has the tension of the enclosed air been reduced ? XIX. Pumps for Liquids, % 125. Common or Suction Pump.— This is an instrument for drawing water from a well or subter- ranean reservoir. It consists of a cylinder fitted with a piston and valve, and connected by a second valve with a pipe which communicates with the reservoir. The cylin- der M N is called the pump-barrel ; the tube v e the suction-tube. The mode of action is as follows : — Suppose the piston at first to be at n, and the suction-tube filled with air at the atmospheric pres- sure. If the piston be raised the air in n d will ex- pand, open the valve v, and follow the piston. At the same time, since the pressure on the surface of the water d by Google The Cofftmon Pump. 153 Fig. 74. Within the tube is diminished, ovAng to the expansion of the contained air, the external pressure at d will cause the water to rise in e to such a height that the elasticity of the air below A B, together with the weight of the column of water above D in the suction-tube^ equals the ainiospheric pressure without. As the piston descends the air below it is compressed, and escapes after a time through the valve f, the valve v being closed by the increased pressure of the air above it. Thus the water remains at the same level in the suction- tube whilst the piston is descending. When the piston is again raised the pressure is removed from above v, and the air underneath it at once opens the valve and occupies the space beneath A B, the water rising in the suction-tube as before. This action continues till the water has risen to n, when the raising of the piston causes the water to enter the barrel, provided the height d n is less than Ji^y the height of the water barometer. As the piston continues to rise the water will follow it so long as the height of the piston above the level of the water outside is less than H, As the piston descends, the pressure of the water beneath it opens the valve f, and the piston passes through the water. When the piston again ascends the water is discharged at the spout, and the barrel is refilled through the suction-tube. The water having once entered the barrel, the Digitized by VjOOQ IC 154 Pnmmatics. contents of the barrel are discharged at each upward stroke of the piston \ but, in order that a volume of water equal to that of the barrel may be discharged at each upward stroke, it is necessary that the atmos- pheric pressure should first raise the water to the spout ; — i,e, the height of the spout above the level of the water outside must be less than H, It is e\4dent that if the piston in descending does not reach v, so as ultimately to exclude all air from underneath it, the water may never be able to enter the pump-barrel, even if n d is less than H, It is not essential to the working of the pump that the tube should be straight ; nor does it matter at what horizontal distance from the pump-barrel the suction-tube enters the reservoir. § 126. Force required to raise the piston-rod. First Suppose the water has not yet entered the barrel. In this case the force necessary to raise the piston-rod is equal to the difference of the pressure of the air on the top and bottom of the piston a b (fig. 74). Let E represent the pressure of the air be- low A B measured in water, and let ZTbe the height of the water barometer outside. Let z equal the height of the column of water in the tube. Then H^=^ E •\- z\ and the force required to mise the piston is A B X {H— -^) = A B v. z = the weight of a column of water that has a b for base, and the distance of surface-level of water in the pump above the level outside for height. Secondly, Suppose the barrel abeady fiill of water. The force required is, as before, the difference of pres- sure on the two sides of the piston- r- i ^ Digitized by LiOOgle The Common Pump. 155 Let jp = c A (fig. 75) be tlie height of the water above the piston at any instant ; then the pressure on the upper surface of the piston is p^^ (Zr4-^)xAB and the pressure on the lower surface is (ZT— <s;)x A B where z is, as before, the height of the water below the piston above the level of the reservoir. Hence, the difference, or force required is AB X {x->rz\ or the weight of the column of water having A B for base and the difference of the level of the water in the pump and in the reservoir for height. This, therefore, is the measure of the tension of the piston- rod in all cases. When the pump is in full action, discharging at each stroke a volume of water equal to that of the barrel, the tension of the piston-rod is constant § 127. Ezamples. (i.) The length of the suction-tube of a common pump is 12 feet, and the piston when at its lowest point is 2 feet from the fixed valve ; if at the first stroke the water rises 1 1 feet in the tube, find the extreme length of the stroke, supposing the water barometer to stand at 33 feet, and the area of the barrel to be three times that of the tube. If jf be the length of the stroke, and a the area of the tube, the volume originally occupied by the air is 12a + 2 x 3^ = i8<j ; the volume occupied by the air after the first stroke is 3a (Jf + 2) + «=(3*+7) a. ,-. by Boyle's law, -il- «33zJ[i=?. 3^+7 33 3 or% Jf=6|feet. Digitized by VjOOQ IC 1 56 Pneumatics, (2.) The suction-tube of a common'pump is 12 feet long, and the piston, starting from the fixed valve, is raised at the first stroke through 3 feet. If the area of the barrel is four times that of the tube, find the height to which the water will be raised in the tube {H = 34 feet). Let X be the required height, a the area of suction-tube. The volume originally occupied by the air was 12a, After the first stroke the air occupies (12— jr) a + 3 x 4/7 = (24— jr) a ; and the pressure is reduced from 34 to 34— ^ feet. Hence, by Boyle's law. 12 . 5£_f , or jc « 8*2 feet nearly M-x 34 § 128. The Liftmg Pomp. — This is a modification of the common pump, in which the water discharged from the pump-barrel, which is closed at the top, en- ters a pipe furnished with a valve and communicating with the spout. The action is the same as before ; but the water, instead of flowing away through the spout, is lifted into the pipe and prevented from returning by a valve. In this way water can be stored up at any elevation, and can afterwards be made to flow through the spout as required. § 129. The Forcing Pump.— In this pump the piston has no valve. The action is the same as in Fig. 76. the common pump until the water enters the barrel. Then, as the piston is raised the water occupies the space beneath it, and passing through the valve c, rises in the tube c d to the same level as in the barrel. As the piston descends the valve b is closed, and all the water contained in the barrel {^forced up the pipe and prevented Digitized by VjOOQ IC ^ The Forcing Pump. 1 17 from returning by the valve C. In this way water can be forced up to any height consistent with the strength of the material, or can be made to nse m a jet from the upper end of the pipe. No flow, how- ever, occurs during the ascent of the piston. § 130. Forcing Pump, with Air-vessel.— To ob- tain a continuous stream of water from the top of the pipe c D (fig. 76) the water must be first admitted into a strong vessel containing air. The action being the same as before, the piston in descending forces the water into the vessel E F (fig. 77), and compresses the air in the upper part of it. As more water is forced into the vessel the air is further compressed, and the water rises in the tube e d and flows out from d. When the piston is drawn up the flow from d would cease but for the fact that the air in e f, freed from its former pressure, now expands and forces the water up the tube, thus causing an unbroken flow from d. The elasticity of the air in e f will decrease with the escape of water from d, and consequently the pressure of the water in the pipe must never be greater than the excess of the pressure of the air in e f over the ordinary atmospheric pressure. If the height of the pipe is inconsiderable, the continuity of flow can be easily preserved during the ascent of the piston. § 131. The Fire Engine. — This is a double forcing pump connected with an air-chamber. The constancy of flow is obtained not only by the air-vessel, but also by the alternate action of the two pumps. The pistons Digitized by VjOOQ IC 158 Pneumatics. are worked by a lever, so that one ascends while the other descends. Every time the pistons momentarily Fig. 78 Stop, the elasticity of the air in the air-chamber maintains the flow. § 132. Brainah*8 Press. — This is a practical appli- cation of Pascal's principle of the equal transmissi- bility of fluid pressure, and consists of an apparatus very similar to that explained in § 27, with the substi- tution of a forcing-pump for the weighted piston. In fig. 79, A is a platform which supports the substance to be pressed against the strong framework B. c and D are two solid cylinders which serve as pistons, and e and f are two hollow cylinders con- nected by a pipe furnished with a valve v. The vessel f communicates with a reservoir of water by the pipe h, so that d f h constitutes an ordi- nary forcing pump. The piston d is attached to a lever, k l m, and the power is applied at m. When the instrument is in action the vessels e and f are filled with water, and as the piston or plunger d descends, it closes the valve z/ and forces the water into the vessel e, and raises the cylinder c, with its attached platform. Digitized by VjOOQ IC BramaKs Press. 159 Fig. 79. r ^ 4- ^ 1 \ c ^kjee! IP 1 •IE « If Q be the pressure produced by the plunger d in its descent, this pressure is distributed equally over the surface of the cylinder c ; and if A be the area of c, A and a that of d, then the pressure exerted at c is — C. d If we suppose P to be the force that must be applied at M, to produce the pressure Q at l, then it follows from the principle of the lever, that Q\ P \\ MK : LK or, G = M K iTk ' If, therefore, W represent the force with which the cylinder, whose area is A^ is pressed upwards, we have W^- . Q = ^ . ^. P; and if ^ and ^ are the a a L K d by Google i6o Pneufnaiics, radii of the two cylinders respectively, and M K = / and L K = ^, we have F f^ ' c This instrument is very extensively employed where the application of a considerable pressure through a small space is required. The very great pressure to which the water is subjected necessitates great care in the construction of the collars through which the cylinders work, so as to render them per- fectly water-tight. Exercises. XIIL • 1. The spout of a common pump is i6 feet from the surface of the water in the reservoir. The area of the pump-barrel is 72 square inches : find the tension of the piston-rod when the pump is full of water. 2. The specific gravity of mercury is 13*6, and the height of the mercurial barometer is 30 inches. What is the greatest height to which water can be raised by means of the common pump? 3. Find the pressure that can be produced by a Bramah's press if the areas of the pistons are 8 : i, and if a force of 10 lbs. is applied at the end of a lever 2 feet long, and at a distance of 20 ins. from the point where the piston-rod is at- tached to it. 4. If in the lifting pump the area of the barrel is four times as great as that of the pipe, compare the pressures on the two valves at the commencement of the third stroke after the water has entered the pipe. 5. If the area of the barrel of the forcing pump be 10 square inches, and of the pipe into which the water is forced 2 Square inches, find the height to which water can be raised in three com- plete strokes, supposing the piston-range to be I foot Digitized by VjOOQ IC Pneumatic Instruments. 1 6 1 6. Find the force that must be applied to the piston-rod, in the preceding question, at the beginning of the third downward stroke. 7. If the diameter of the piston of a common pum{> be 4 inches, and the height of the head of the water in the pump 18 feet above the well, find what pressure the piston bears, taking the weight of i cubic foot of water equal to 62*3 lbs. 8. The length of the suction-tube is 20 feet, and the entire stroke of the piston is 6 feet The piston starts from the bottom of the barrel, and at the end of the first stroke the water has risen 12 feet in the tube. Compare the area of the barrel with that of the tube {H^ 34 feet). 9. A small strong pump is employed for raising mercury from a vessel. The height of the fixed valve is 2 feet above the level of the mercury in the vessel, and the spout is 8 in. above the valve. When the pump is in full action what part of the con- tents of the barrel can be ejected at each stroke of the piston (^= 30 in. ) ? 10. A suction-tube is used for drawing up mercury from 3 vessel containing it, and the piston is raised 10 inches from its lowest point, which is 2 inches from the valve at the bottom ol the tube. To what height will the mercury rise in the tube {iy=3oin.)? Vi d by Google MISCELLANEOUS PROBLEMS. 1. A CUBE of brass, whose edge is 2 inches and specific gravity 8, is completely imbedded in a cube of wood, whose edge is 3 inches and specific gravity 5. Find the mean specific gravity of the whole cube. 2. When equal volumes of alcohol (specific gravity «= o*8) and distilled water are mixed together, the volume of the mix- ture (after it has returned to its original temperature) is found to fall short of the sum of the volumes of its coastituents by 4 per cent. Find the specific gravity of the mixture. (Univ. Lond. Matric.) 3. The specific gravity of cast copper is 8*88, and that of copper wire is 8*79. "What change of volume does a kilo- gramme of cast copper undergo in being drawn out into wire? (Univ. Lond., istB. Sc.) 4. A mixture is made of 7 cubic centimetres of sulphuric acid (specific gravity = I '843^ and 3 cubic centimetres of dis- tilled water, and its specific gravity when cold is found to be 1*615. Determine the contraction which has taken place. (1st B. Sc. 1874.) 5. Two liquids are mixed (i) by volume in the proportion of I : 4, arid (2) by weight in the proportion of 4 : I. The re- sulting specific gravities are 2 and 3 respectively. Find the specific gravities of the liquids. Digitized by VjOOQ IC Miscellaneous Problems. 163 6. The specific gravity of a mixture of two different liquids being supposed to be an arithmetic mean between those of the component liquids ; required the ratio of the volumes of the latter contained in the mixture. (Univ. Lond., B.A. 1877.) 7. Find the pressure on a vertical rectangle, 10 inches long and 6 inches broad, immersed in water with its longer sides horizontal and with the upper one 2 inches below the surface. (One cubic foot of water weighs i,ooo ounces.) (Matric. 1877.) 8. A vessel in the shape of a pyramid, 5 feet high, and with a base 4 feet square, is filled with water. Find the pressure upon the base, and account for its being greater than the total weight in the vessel. 9. Find the whole pressure on the lower half of the curved surface of a vertical cylinder filled with water, the area of the base being I2| square cm. and the height 1*4 decim. 10. A piston, 6 square inches in area, is inserted into one side of a closed cubical vessel, measuring 10 feet each way, filled with water : the piston is pressed inwards with a force of 12 lbs. Find the increase of pressure produced on the entire surface of the vessel. (B. A. 1872.) 11. The pressure at the bottom of a well is four times that at the depth of 2 feet ; what is the depth of the well if the pressure of the atmosphere is equivalent to 30 feet of water ? (Camb. Gen. Exam. 1877.) 12. A and B are vessels fiiU of water, with circular and horizontal bases, 12 inches and 8 inches in diameter respec- tively. A is 8 inches, and B is 9 inches high. Compare the pressure on the bases. (Cam. Gen. Exam.) 13. If a piece of wood weighing 120 lbs. floats in water with four-fifths of its volume immersed, show what is its whole volume (I cubic foot of water weighs 625 lbs.). (Matric. 1872.) 14. A wine-bottle, which below the neck is perfectly M 2 d by Google 164 Hydrostatics and Pneumatics, cylindrical and has a flat bottom, Is placed in pure water. It is found to float upright, with 4J inches immersed. The bottle is now removed from the wj^ter and put into oil, the specific gravity of which is 0*915. How much of it will be immersed in the latter fluid? (Matric. 1874.) 15. Two pieces of iron (specific gravity 77) suspended from the two scale-pans of a balance, the one in water and the other in alcohol of the specific gravity 0*85, are found to weigh exactly alike. Find the proportion between their true weights. (B. Sc. 1875.) 16. An inch cube of a substance of specific gravity i '2 is immersed in a vessel containing two fluids which do not mix The specific gravities of these fluids are i*o and i -5. Find what will be the point at which the solid will rest. (Matric. 1876.) 17. A substance which weighs 14 lbs. in air and 12 lbs in water, floats in mercury whose density is 13*6. What propor- tion of its volume will be immersed ? 18. A solid, of which the volume is 1*6 cubic centimetres, weighs 3*4 gprams in a fluid of specific gravity 0*85. Find the specificgravity and weight of the substance. (B. Sc 1876). 19. An accurate balance is totally immersed in a vessel of water. In one scale-pan some glass (specific gravity 2*5) is being weighed, and exactly balances a one-pound weight (specific gravity 8*o), which is placed in the other scale-pan. Find the real weight of the glass. (Matric. 1875. ) 20. A right cone, whose weight is W, floats in a liquid, vortex downwards, with \ of its axis immersed ; what addi- tional weight must be placed on the base of the cone so as just to sink it entirely in the liquid. (Woolwich Exam.) 21. A cube floats in distilled water under the pressure of the atmosphere, with four-fifthsof its volume immersed and with two of its faces horizontal. When it is placed under a con- denser where the pressure is that of ten atmospheres, find the Digitized by VjOOQ IC Miscellaneous Problems. 165 alteration in the depth of immersion (the specific gravity of air at the atmospheric pressure being; •cx)i3). (B. Sc.) 22. At the bottom of a mine a mercurial barometer stands at 77*4 centimetres ; what would be the height of an oil baro- meter at the same place, the specific gravity of mercury being 13-596, and that of oil 09 ? (Matric. 1876.) 23. A certain quantity of air at atmospheric pressure has a volume of 2 cubic feet, the temperature being 55° Fahr. What does the volume of the air become when the pressure is increased by one-twentieth, the temperature meanwhile remaining the same? (Women*s Exam. 1876.) 24. A syphon barometer is so constructed that the long closed tube has an internal sectional area equal to \ of an inch, while the short open tube has an internal sectional area equal to J an inch. Find what fall will take place in the long tube of this barometer when the true pressure of the air falls one inch. (B. Sc. 1875.) 25. The mercury in a barometer stands at £o inches ; the section of the tube measures I square inch, and "the vacuum above the mercury 6 cubic inches, as much air is passed up the tube as depresses the mercury to 29 inches : what would be the space occupied by the air under the atmospheric pressure ? (B. Sc. 1872.) 26. In a tube of uniform bore a quantity of air is enclosed. What will be the length of this column of air under a pressure of three atmospheres, and what under a pressure of a third of an atmosphere, its length under the pressure of a single atmosphere being 12 inches? (B. A. 1876.) 27. A Marriotte's tube (fig. 51) has a uniform section of i square inch, and is graduated in inches, 6 cubic inches are enclosed in the shorter (closed) limb, when the mercury is at the same level in both tubes. What volume of mercury must be poured into the longer limb, in order to compress the air Digitized by VjOOQ IC 1 66 Hydrostatics and Pneumatics. into 2 inches ? The barometer stands at 30 inches. (B. Sc. 1874.) 28. Two cubic centimetres of air are measured off at atmo- spheric pressure. When introduced into the vacuum of a baro- meter they depress the mercury which previously stood at 76 cm., and occupy a volume of 15 cubic centimetres. By how much has the mercurial column been depressed? (Matric. 1878.) 29. A tumbler full of air is placed mouth downwards under water, at such a depth that the surface of the water inside it is at a depth of 25^^ feet. Compare the weight of a cubic inch of air in the tumbler with that of a cubic inch of air outside — the barometer standing at 30 inches, and the specific gravity of mercury being 13-6. 30. The contents of the receiver of an air-pump is six times that of the barrel. Find the elastic force of the air in the re- ceiver at the end of the eighth stroke of the piston, when the atmospheric pressure is 15 lbs. to the square inch. (B. A. 1872.) Richard Clay ^ Sons, Limited^ London &• Bungay, Digitized by VjOOQ IC d by Google d by Google d by Google I'Jl ^ J^ /; l-^'->'7 '>-'^ / i b d by Google I d by Google